GEOMETRICAL OPTICS
GEOMETRICAL OPTICS
BY
ARCHIBALD STANLEY PERCIVAL
M.A., M.B., B.C. CANTAB.
SENIOR SURGEON NORTHUMBERLAND AND DURHAM EYE INFIRMARY
AUTHOR OF "OPTICS," "PRACTICAL INTEGRATION," "PRESCRIBING OF SPECTACLES 1
WITH
LONGMANS, GREEN, AND CO,
39 PATERNOSTER ROW, LONDON
NEW YORK, BOMBAY, AND CALCUTTA
1913
[All rights re se
PREFACE
THIS book is primarily intended for medical students as a
textbook on the subject of Geometrical Optics for their
preliminary scientific examinations, though it practically
contains all the Optics required by an ophthalmic surgeon.
It is hoped that it will also prove of service to students of
physics, as some knowledge of the subject is indispensable
if the laboratory experiments are to be understood.
It requires prolonged and deep study to form any vivid
conception of the now accepted theory of light, and in all
elementary books the form in which the undulatory theory
is presented is so crude that it is both untrue and useless.
The subject of Physical Optics therefore has been avoided
entirely; indeed, I am convinced that no thorough elemen
tary knowledge of that intricate subject can be obtained in
the short space of time allotted to the student for studying
Optics.
As an introduction to mathematical analysis the subject
of Geometrical Optics has no equal, for it insists on the
importance of paying due attention to the meaning of
algebraic signs, and it is also an easy introduction to several
somewhat difficult mathematical conceptions. For instance,
the vectorial significance of the line BA being considered as
equal to AB, or AB taken in the reverse direction, opens
up a new vista to the student of Euclid and elementary
geometrical methods: equally novel is the conception of a
virtual image. At the same time every student can verify
for himself the results of his calculations so simply by
experiment that it will convince him of the reality of the
analytical methods employed.
263251
vi PEEFACE
I have embodied in the text all that can be reasonably
demanded in any preliminary examination in science, while
the Appendix contains matter that will be of service sub
sequently in a professional career. The starred paragraphs
may be omitted at a first reading, as a knowledge of them
would be rarely required in the examinationroom. The
subject of Cardinal Points is always neglected in the elemen
tary books ; it has here been treated, I venture to think, in
a much simpler manner than is customary in the solution
of the problem; for the ingenious graphic solution on
p. 106 I am indebted to Professor Sampson. The subject is
of first importance in understanding the optical properties
of the eye, and it is of the greatest value in dealing simply
and readily with many otherwise difficult questions.
The reader will find a list of all the important formulae
for ready reference at the end of the book, and he will notice
that some of them are of universal application, and that by a
simple transformation, formulae for refraction can be converted
into the corresponding formulae for reflection. The proofs
while preserving their rigid character are made as simple as
possible, and the utmost care has been taken to include only
what is of practical application, excluding all that is of
merely academic interest.
ARCHIBALD STANLEY PERCIVAL.
17, CLAREMONT PLACE,
NEWCASTLEUPONTYNE,
Dec. 1912.
CONTENTS
PAGE
I. ILLUMINATION PINHOLES SHADOWS 1
IL REFLECTION AT PLANE SURFACES 11
III. REFLECTION AT A SPHEBICAL SURFACE 18
IV. REFRACTION AT PLANE SURFACES 34
V. REFRACTION AT A SPHERICAL SURFACE 49
VI. LENSES 68
APPENDIX 109
ANSWERS 129
INDEX 131
CHAPTEE I
ILLUMINATION PINHOLES SHADOWS
WE shall here confine ourselves to the study of some of the
simplest properties and laws of Light. When we say that
we see the sun or a tree we merely mean that we see the
light that comes from them ; the sun, of course, sends out
light of its own, whereas the tree passes on or reflects the
light chat it receives from something else. It is obvious
that in neither case do we see the thing itself ; we are only
conscious of a certain senseimpression derived from it. The
nature of this senseimpression and the way in which it is
developed from a physical stimulation of the retina, are
problems that are still engaging the attention of physio
logists and psychologists ; with questions such as these the
science of Optics does not deal.
We shall commence our study by the consideration of the
following laws :
LAW 1. In a homogeneous medium light is propagated in
every direction in straight lines.
LAW 2. The intensity of illumination varies inversely as
the square of the distance from the source of
light, and it is greatest when the angle of
incidence is 0.
LAW 3. When the incident rays of light are parallel, the
intensity of illumination varies as the cosine
of the angle of incidence.
Our experience of shadows confirms the truth of Law 1.
A luminous point in space sends out light in all directions,
2 GEOMETEICAL OPTICS
illuminating a rapidly enlarging sphere. It is the radii of
this sphere that are called rays of light ; they are simply the
directions in which the light is travelling, and when this
direction is altered by reflection or refraction the rays con
sidered are bent at an angle. But for either reflection or
refraction to occur a heterogeneous medium must be
encountered; so that we conclude that in a homogeneous
medium light is propagated in every direction in straight
lines.
Experiments with diffraction gratings show us that Law 1 is not
strictly true, and, had we to treat of the phenomena of Diffraction and
Polarization, we should have to explain in detail the electromagnetic
theory of light, or at any rate to give an account of some form of wave
theory. As these subjects do not now concern us, we may regard the
directions in which light travels as straight lines or rays, and in this
way the elementary study of optics will be much simplified.
The second and third laws will require some careful
explanation and consideration before they will be accepted
by the reader.
Illumination. Let S be a source of light, and consider
the light that is being propagated from S in the direction of
the screens AK and BL (Fig. 1). It will be noticed that,
FIG. i.
since light travels in straight lines, the amount of light (Q)
that falls upon the square AK is the same as that which
would illuminate the larger square BL if AK were removed.
Now, in the diagram SB is made equal to 3 SA, so the area
of the square BL is nine times the area of AK. Hence the
ILLUMINATION 3
illumination per unit area, or the intensity of illumination, of
BL is of that of AK.
Indeed, B T : A= 1:9 "
It is obvious, then, that the intensity of illumination
varies inversely as the square of the distance from the source
of light.
Nearly all photometers are based upon this law ; we will
describe that designed by Eumford. Suppose that we wish
to compare the intensities, I and L, of illumination of two
lamps ; these are placed at distances d and D from a white
screen so that two sharp shadows of a vertical rod placed a
few inches in front of the screen are thrown upon it ; each
shadow is then illuminated by only one of the lamps. Thus
if L, the standard light, be regarded as 1, the other lamp is
moved backwards or forwards until the position d is found
in which both shadows appear equally dark. Each lamp is
then sending an equal amount of light to the screen, and
the relative illuminating power is given by the ratio of
the squares of their respective distances (d? and D 2 ) from
the screen. The light that the two shadows receive are
respectively ^ and^; consequently when these are equal,
1 : L = d 2 : D 2 . Thus if D the distance of the standard
light be 2 feet, while d = 4 feet,
1 : 1 = d? : D 2 = 16 : 4 = 4 : 1
or I is four times the intensity of the standard light.
It is commonly said that four candles at a distance of
2 feet give the same illumination as one candle at a
distance of 1 foot ; this, although true, could not be satis
factorily proved by a photometer which measures only the
intensity of illumination from a point source of light.
It may be noted that the accuracy with which such tests
can be made depends upon the " Light Sense " of the observer,
so that if the values of I and L be known, this photometer
may be used as a test of the observer's " Light Difference."
On referring to Fig. 1, it will be noticed that both AK
4 GEOMETKICAL OPTICS
and BL are at right angles to SB. Now, the angle of incidence
(0) is the angle which a line at right angles to the surface
of the screen makes with the incident ray ; in this case, there
fore, the angle of incidence is 0. If the screen were inclined
either forwards or backwards, turning round its base line, it
is clear that the intensity of illumination per unit area of
the screen would be diminished, for in either case some of
the cone of light that is represented would not fall upon the
tilted screen. The diminution of illumination approximately
follows the cosine law (see below) when the incident cone
of light is not too divergent.
If the illuminant were a beam of parallel rays, or were at
an infinite distance as compared with the distance AB, the
law of inverse squares ceases to have any intelligible meaning ;
for then both SA and SB are infinite, and the illumination of
a surface exposed to such a light is constant whatever its
distance. For instance, the light from a morning sun on a
screen is practically the same as the light on a similar screen
placed at a mile's distance to the west.
If I be the intensity of illumination a screen receives from
a beam of parallel rays when the angle of incidence is 0, it
is clear that when the screen is tilted so that the angle of
incidence is 0, the illumination is I cos 0.
Apparent Brightness. The distinction between brightness
and illumination is hardly made clear enough in the books.
Illumination refers to the physical condition of the object
when illuminated, whereas brightness refers to the resulting
senseimpression produced in the observer. It will clearly
depend upon his physiological condition ; but, neglecting this
for the moment, let us consider how an illuminated screen at
BL will appear to an observer at S;
The brightness of an object is naturally measured by the
amount of light it sends to the eye per unit area of its
apparent size; in other words, the brightness (B) of an
object is directly proportional to the quantity of light (Q)
that it sends to the pupil, and inversely proportional to the
apparent area (A) of the surface observed, or B = ~.
A
APPARENT BRIGHTNESS 5
Since, however, both Q and A are each of them inversely
proportional to the square of the distance of the luminous
object, the apparent brightness is independent both of its
size and of its distance, presuming of course that the medium
is clear and no adventitious absorption occurs in it. If the
physiological conditions be the same, the apparent brightness
of an object merely depends upon the intrinsic luminosity of
the object. A redhot iron ball is indistinguishable from a
circular disc of iron at the same temperature, showing that
the brightness is independent of the inclination of the
periphery of the ball to the line of sight. The sun and
moon may both be regarded as approximately spherical, yet
both appear to the naked eye as flat discs of uniform bright
ness. There are, however, several physiological conditions
that greatly affect the apparent brightness of luminous
bodies.
(1) The size of the pupil. The brightness (B) must vary
as the area of the pupil. The brightness of an object will
be much diminished by viewing it through a small pinhole,
for in that case less light enters the eye, as the size of the
pupil is virtually diminished.
(2) The condition of retinal adaptation. Prolonged stay
in a dark room will much increase the apparent brightness
beyond what is due to the consequent dilatation of the pupil.
When high powers of a microscope or a telescope are
used, the brightness of the image is much diminished, as
then only part of the pupil is filled with light. This is
practically equivalent to lessening the area of the pupil (1).
It is impossible by any optical arrangement to obtain an
image whose brightest part shall exceed the brightest part
of the object. If allowance be made for the loss of a certain
amount (about 15 %) of light by reflection and imperfect
transparency of the lenses, the brightness of the image is
equal to the brightness of the object.
There is one case in which this law does not hold good,
and this also depends upon physiological reasons. If the
object be extremely small, it may yet if excessively bright
succeed in stimulating a retinal cone, and so cause a visual
6 GEOMETKICAL OPTICS
impression of a tiny point of light, even although its image
does not cover the whole surface of the cone. If such an
object be magnified until its image covers the cone, there will
be no increase of its apparent size although an increased
amount of light will be entering the eye. When, for
instance, stars are observed through a telescope, their
apparent size is not increased, for their image does not
extend beyond one retinal cone, but all the light that falls on
the objectglass may by a suitable eyepiece be concentrated
on the observer's pupil, except that lost by transmission
through the lenses. If, then, a denote the fraction of the
incident light that is transmitted through the telescope
(about 0*85), and O denote the area of the objectglass, and e
that of the pupil, the increase of brightness will be a . Or
Q
if o and p be the diameters of the objectglass and the pupil
o
respectively, the increase of brightness will be a 3 , for the
areas of circles are proportional to the squares of their
diameters. If the pupil be regarded as of unit diameter we
get the expression ad 2 . This is what astronomers call the
" space penetrating power " of a telescope, that is its power of
rendering very small stars visible.
Precisely the same conditions obtain with the ultrami
croscope, which is a device for bringing into view fine
granules that are smaller than the limit of the resolving
power of the instrument. By means of darkground illumi
nation these fine particles are relatively so brightly lighted
that they succeed in stimulating a retinal cone even though
their image does not cover its whole surface. ^Resolution
does not occur; whatever its real shape the granule will
appear as a circular point of light, so that all one can really
say is that something very small is present.
Visual Angle. On referring again to Fig. 1 it will be
seen that if the observer's eye were situated at S the small
square AK would completely obliterate the larger square BL,
as they both subtend the same angle at S. Such an angle is
called the visual angle, and it is obvious that if one does not
PINHOLES 7
know the distance of an object one can form no real idea of
its size. The expression " apparent size " is quite illusory ; it
conveys no clear meaning. For instance, some will say that
the sun's apparent size is that of a dinner plate, others that
of a saucer, but it will be found on trial that a threepenny
bit at a distance of 6 feet will completely obliterate it, yet
few would assert that its apparent size was that of a three
penny bit even when held at arm's length. The importance
of the visual angle will appear when we deal with magnifi
cation.
Pinholes. If a pinhole be made in a card, and this be
held between a candle and a screen, an inverted image of the
candle flame will be formed upon the screen. The nearer
the screen is brought to the pinhole the smaller and sharper
will be the image. If the candle be brought nearer the pin
hole, the image will be larger but less sharp. Finally, if the
hole in the card be made larger the image will appear
brighter, but its definition will be again diminished.
The explanation is simple, as will appear from an exami
nation of Fig. 2. Every point of the candle flame is sending
out light in all directions, and all that falls upon the card is
intercepted by it, so that its shadow is thrown upon the
screen. From each point of the candle flame, however, there
will be one tiny cone of light that will make its way through
the pinhole. On the screen the section of this cone will
appear as a small patch of light. Thus the point A of the
flame will be represented by the patch a, and the point B by
the patch & on the screen. Clearly the height ab of this
inverted image will be proportional to the distance of the
screen from the pinhole ; if AB be brought nearer the card
each pencil will form a more divergent cone, or if the hole be
made larger the same result will occur ; hence in both these
cases the patches a and & will be larger and the definition
will be impaired.
If care be taken that these patches of light are not too
large, an image sufficiently sharp for photographic purposes
may be obtained. For some purposes a pinhole camera made
out of a preserved meat tin may prove a more satisfactory
8
GEOMETRICAL OPTICS
instrument than one provided with a lens. The disad
vantage of a pinhole camera is of course the feeble illumi
nation of the image, which makes a very long exposure
necessary.
FIG. 2.
The best size of pinholes for photographic work, according to
Abney, is determined by the approximate formula
p = 0008 Jd
Thus, if the distance between the pinhole and the plate be 9 ins., the
diameter of the pinhole should be O'OOS x 3 = 0'024 in.
Shadows. When the source of illumination may be
regarded as a point, it is clear that an opaque obstacle
like AK in Fig. 1 will be lighted only on its proximal sur
SHADOWS
9
face, while the prolongation of the cone of light indicates the
cone of shadow cast by AK.
If, however, the source of light be a luminous body
possessing innumerable points, the object will be illuminated
by innumerable cones of light, and we must imagine a shadow
for each of them behind the object. The space behind the
object which is common to all of these shadow cones will
represent the area of total shadow, or umbra. There will,
however, be a space outside this which is only in shadow as
regards part of the luminous body while it receives light from
another part of it, and is consequently partially illuminated.
This is the area of partshadow, or penumbra. In the adjoin
ing illustration (Fig. 3) two opaque bodies are represented,
FIG. 3.
one being smaller and the other larger than the luminous
body which is placed between them. In each case for the
sake of clearness the limits between umbra and penumbra
have been more sharply defined than they should be. When
a total eclipse of the sun occurs, the moon is so situated that
it intercepts all the light coming from the sun towards the
earth ; the earth is then in the umtoa : when a partial eclipse
occurs, the earth is in the penumbra.
10 GEOMETRICAL OPTICS
QUESTIONS.
(1) What is the height of a tower that casts a shadow 52 feet 6
inches in length on the ground, the shadow of a stick 3 feet high being
at the same time 3 feet 6 inches long ?
(2) A pinhole camera, the length of which is 7 inches, forms an
inverted image 4 inches high of a house that is in reality 40 feet high.
What is the distance of the camera from the house ?
(3) A gas lamp distant 5 feet and an electric light distant 150 feet
throw on an opposite wall two shadows of a neighbouring post. If
these two shadows are of equal intensity, what is the relative illuminat
ing power of the lights ?
(4) If the electric light in (3) were raised vertically to such a height
that its distance from the wall were 300 feet, what would be approxi
mately the relative intensities of the shadows ?
CHAPTER II
REFLECTION AT PLANE SURFACES
WE must now consider the manner in which light is reflected
by polished surfaces. The following two laws when properly
understood explain every possible case of reflection :
LAW 1. The reflected ray lies in the plane of incidence.
LAW 2. The angles which the normal (at the point of
incidence) makes with the incident and with
the reflected ray are numerically equal, but
they are of opposite sign.
These laws hold good whether the surface be plane or
curved ; in the latter case we have only to draw a tangent
at the point of incidence, and consider the ray reflected at
this plane. The meaning of the technical terms and the
sign of an angle will be explained in the next paragraph.
Reflection at Plane Surfaces. Let AB represent a plane
reflecting surface (Fig. 4), and let S be a luminous point
sending out light in all directions. Now, if SN be one of
these directions, SN represents an incident ray, N the point
of incidence, and NY the normal at that point. Now, the
plane of incidence is that plane that contains both the
incident ray and the normal, so the plane of the paper is the
plane of incidence. According to the laws just stated NQ is
the corresponding reflected ray, for the angle of incidence
YNS is numerically equal to the angle of refraction YNQ,
which lies in the plane of the paper, and is of opposite sign to
YNS. The angle YNS is measured by the rotation at N of
a line in the direction NY to the position NS, i.e. in the
direction of the hands of a clock, whereas YNQ is measured
11
12
GEOMETEIOAL OPTICS
in the counterclockwise direction. This is what is meant by
the sign of an angle, so that if YNQ be denoted by $' and
YNS by 0, we have in reflection 0' =  <.
If, then, there be an eye in the neighbourhood of Q it will
receive light coming towards it in the direction NQ ; but we
have not yet found from which point in this line it will appear
to have come. To do this we shall have to take another
incident ray, SM say, and discover where the corresponding
reflected ray intersects the previous one.
FIG. 4.
It is clear that as the incident and reflected rays make
equal angles with the normal they must make equal angles
with the mirror ; so we made the angle BMR numerically
equal to AMS, and produce the reflected ray RM to meet QN
produced in S'. Then S' is the point from which the light will
appear to have come.
Now, it is clear that in the triangles SMN and S'MN we
have the base MN common, and the angles at the base equal,
so the triangles must be equal ; and if we join SS', we see that
the two sides SM, MA are equal to the two sides S'M, MA,
and the included angle SMA is equal to S'MA ; so S'A is
equal to SA and the line SS' is at right angles to AB.
Similarly, it may be shown that any other ray in the same
CONSTRUCTION OF THE IMAGE
13
plane will be reflected in such a direction that when produced
backwards it will meet SS' in S'. Now, if we suppose the
paper to be revolved about an axis SS', the figure will repre
sent the course of incident and reflected rays in every plane.
Hence all the rays that fall upon the mirror from S, whatever
the plane of their incidence, will be so reflected that the
prolongations of these reflected rays will intersect at the
point S', so that an image of S will be formed at S'.
It will be noted that this image has no real existence; the
reflected rays do not really
come from the back of the
mirror, they only appear
to come from the point S' :
such an image is called a
virtual image. Subse
quently we shall deal with
real images, and the dis
tinction between them is
simply this : Real images
are formed "by the inter
sections of the reflected (or
refracted) rays themselves ;
virtual images are only
formed by the intersections
of their prolongations.
Construction of the
Image of an Object. Sup
pose that it is required
to construct the image of the object Pp as seen in the
mirror NM (Fig. 5); all we have to do is to draw lines
perpendicular to the mirror (or its prolongation) from P
and p to Q and q, so that Q and q are as far behind the
mirror as P and p are in front of it ; then Qq is the virtual
image of Pp. If an eye be situated in the neighbourhood of
E, and we wish to show the actual course of the light that
reaches the eye from P, we join EQ, cutting the mirror at M,
and then we join MP. The light that forms the image Q for
an eye at E reaches it by the path PM and ME, Another
FIG. 5.
14 GEOMETEICAL OPTICS
eye at K' will see the same image Q by means of light that
travels in the direction PN and NE'.
It will be noted that the image Qq is similar and equal
to the object "Pp, for they both are similarly situated with
respect to the mirror. Since, however, the object and the
image face each other, the observer gets a view of that side of
the object that he could not see without turning it round.
Hence he thinks the object has been turned round so that its
right and left sides have been interchanged. For this reason
the image of a printed letter will appear " perverted," that
is, it will appear upright, but it will resemble the type from
which it was printed.
Deviation produced by Rotation of Mirror. If the
mirror AB (Fig. 6) be slowly rotated into the position cib,
the image S' will appear to rotate through an angle twice
as great in the same direction, so that finally it will appear
to have moved to S". The explanation is obvious. If S be
the object, when the mirror is in the position AB its image
will be at S' ; on rotating the mirror through an angle 9 its
normal will also rotate through the same angle, so that
SIY = 9, but then reflection will occur in the direction
IK, and the image of S will appear at S". Clearly,
SIS" = SIE = 29.
There are several important applications of this principle
in daily use ; as examples we may mention the sextant and
the mirror galvanometer, but we will describe in greater
detail the laryngoscope. When this instrument is used, the
mirror, which is inclined at an angle of 45, is placed at the
EEPEATED KEFLECTION
15
back of the pharynx; when properly placed an image of
the larynx is seen with its axis horizontal. Since the image
and object are similarly situated with respect to the mirror
and not to the observer, the anterior parts of the larynx
(epiglottis, etc.) are represented in the upper part of the
image, while the posterior structures (arytenoids, etc.) occupy
the lower portion of the image.
Repeated Reflection at Inclined Mirrors. When an object
is placed between two plane mirrors inclined at an angle, a
limited number of images may be seen by an observer in a
FIG. 7.
suitable position. If a be the angle between the mirrors,
,.180 ,
and it = 7i, the number of images seen by an eye in a
suitable position will be 2n  1, if n be an integer. Moreover,
the object together with its images then forms a perfectly
symmetrical figure with respect to the reflecting surfaces.
Fig. 7 represents two mirrors, AC and BC, inclined at an
angle of 60 with an object PQ between them. As 2n 1 in
this case is 5, 5 images are seen, and the symmetry of the
figure is evident. This is the principle of the kaleidoscope.
16
GEOMETEICAL OPTICS
1 80
Suppose that a = 40, then n = ^ or 4 J ; in such a case
it will depend entirely on the position of the observer
whether 4 or 5 images are seen.
The position of the inages is easily found by the construction now
familiar to the reader. Let AC, BC (Fig. 8) be two plane mirrors
inclined at an angle of 90, and let^? denote the position of an object
between them. From p draw perpendiculars to each mirror, and
produce them to pf and p^ so that pf and p* are as far behind the
FIG. 8
mirrors AC and BC respectively asjp is in front of them. Then pf and
pj are two of the images. To find the position of the third image p n ,
we must produce AC and BC and consider these prolongations as
mirrors and the images we have already found as objects. That is
to say, we make p u either at an equal distance behind AC produced
to that of p^ in front of it, or we make p n an equal distance
behind BC produced to that of pf in front of it. It will be noticed
that whichever construction is used, p n occupies the same place. To
find the path of light from p n to an eye at E, join jp n E, cutting BC
in Q ; join pfQ, cutting AC in K ; join jpR. Then the path of the light
from p is ^B, RQ, QE. If E had been close to AC, on joining _p n E
the mirror AC instead of BC would have been cut, so that in such a
case the last reflection would have been from the mirror AC,
EEPEATED REFLECTION 17
must be joined to the intersection on AC. These constructions are of
little importance except for examination purposes.
When the mirrors are parallel to each other, the angle
1 80
between them is 0, and since ^ = oo , an infinite number
of images might be seen if certain physical conditions did
not prevent their observation.
QUESTIONS.
(1) Two parallel mirrors face each other at a distance of 3 feet,
and a small object is placed between them at a distance of 1 foot from
one of them. Calculate the distances from this mirror of the three
nearest images that are seen in it.
(2) Two mirrors, AC and BC, are inclined at an angle of 45 ; an
object, P, is so placed that ACP = 15. How many images will an
observer halfway between the mirrors see ? Trace the path of light
that gives rise to the second image p n b seen in BC.
(3) If in the last example BC were rotated so that ACB = 135,
how many images would an observer (E) halfway between the
mirrors see ? In what position would he see more ?
CHAPTEK III
REFLECTION AT A SPHERICAL SURFACE
EEFLECTION at irregularly curved surfaces can only be treated
by the method mentioned on p. 11. At the point of
incidence a tangent plane is drawn, and for that particular
incident ray the reflection is considered as taking place at
that tangent plane. Fortunately, however, there are much
less tedious methods of dealing with reflection at spherical
surfaces, with which alone we are now concerned ; these we
proceed to describe in their simplest form.
Concave Spherical Mirrors ; Axial or Centric Pencils. Let
AK represent a concave mirror, its surface being part of a
sphere of which the centre is at C (Fig. 9). Any line drawn
through C to the mirror is called an axis of the mirror, and
when the parts of the mirror on either side of the axis are
symmetrical it is called the Principal Axis. In the figure
CA is the principal axis, and A the vertex of the mirror.
Let P be a luminous point on the axis distant PA or p from
the mirror, and consider the incident ray PK when the point
K is near the vertex A. Join CK and make the angle CKQ
numerically equal to the angle of incidence CKP. If now
the figure be supposed to revolve round the principal axis
PA, it is clear that PK will trace out the limits of a thin
axial cone of incident light, while the point K will trace
out a narrow circular zone on the mirror, and the reflected
rays from this zone must all intersect the axis at the point
Q. Consequently Q must be the image of P as formed by
reflection at this narrow zone. Let CA, the radius of the
mirror, be denoted by r, and let QA be denoted by q.
18
CONCAVE SPHERICAL MIREOKS
19
Now, since in the triangle PKQ the vertical angle is
bisected by KC, which cuts the base at C
(Euc. vi. 3)
QK ~~ CQ ~" CA  QA rq
When K is very near the vertex A, PK and QK will be
almost identical with PA and QA; under these circum
FIG. 9.
stances we may replace PK and QK by p and q, and so we
get the following formula for a very thin centric pencil :
P = pr
.'. pr pq = pq rq or qr + pr = 2pq
On dividing by pqr we obtain the formula
pqr
The point Q marks the situation of the real image of P ;
it is a real image because it is formed by the intersection
of the reflected rays themselves (see p. 13). Again, Q is
often called the conjugate focus of P. It is termed a focus
(fireplace) because the place where heat rays intersect is
hotter than any other. If, for instance, the mirror be used
to form an image of the sun, a piece of paper placed at the
20 GEOMETKICAL OPTICS
focus will be burnt up. The term "conjugate" implies that
if the object P be placed at Q, the image will then be in the
old situation of P. In fact, P and Q are mutually con
vertible whenever the image formed at Q is real.
We will now consider the formula a little more closely.
If PA or p be diminished until p = r y QA or q becomes
greater until q becomes equal to r, for
q r p r r r
This only means that when the luminous point is placed
at C all the incident light reaches the mirror in the direction
of its radii, or the normals to the surface, so that the
reflected light must travel back by the same path ; the image
Q is then coincident with the object at C.
If the object P be brought still closer to the mirror
(between C and F), the image Q will be formed at a greater
distance from it ; in fact, the situations of P and Q will be
interchanged.
Now let us suppose the object P to be removed to a great
distance ; the image Q will be formed nearer the mirror. In
the limiting case when the distance of P is infinite, let us
see what our formula tells us. Here p = oo .
so +  = . But 1 = 0, so  = 
cc q r oo q r
This means that if the object is at an infinite distance, or,
what comes to the same thing, if the incident beam consist
of parallel rays, the corresponding focus is at a point F such
that FA = CA. This point F is called the Principal Focus,
and the distance FA is usually symbolized by/. Clearly,
as the direction of light is reversible, it follows that if a
luminous point be placed at F the reflected rays will form a
beam of light parallel to the axis.
T
Since we now know that / = ~, we may give the previous
formula in its more usual form
M = U/ + /=i
P 2 / P 2
SIGNS 21
What will happen if the object be placed nearer the
mirror than F? Our formula will tell us. Suppose that
/ = 4, and that p = 3 ;
QA or q is then 12. What is the meaning of the
negative sign ? We have regarded the direction from left to
right (e.g. PA and CA) as positive ; the opposite direction,
from right to left, must therefore be the negative direction.
Consequently, QA must now be measured in this direction,
that is, the point Q must be behind the mirror. The reader
is urged to draw for himself a diagram illustrating this case ;
that is, let him draw an arc of 8 cm. radius with its focus at
4 cm. from the mirror, and from a point P, 3 cm. from the
mirror, let him draw any incident ray, PK, to the mirror ;
let him then mark a point Q, 12 cm. to the right of (behind)
the mirror, and join QK and produce it. He will find that
the normal CK bisects the angle between PK and the pro
longation of QK, showing that KQ is the prolongation of the
reflected ray in this case in other words, Q is a virtual
image of P, as it is formed, not by the reflected ray itself, but
by its prolongation (see p. 13).
Whenever Q is a virtual image, P and Q are no longer
interchangeable ; in this case, for instance, the object if placed
in the situation of Q would be behind the reflecting surface
of the mirror, so no image would be formed.
Signs. This case will indicate the importance to be
attached to the meaning of algebraic signs. It will be found,
if due attention be paid to them when thin axial pencils
are being considered, that there are only two formulae that
need be remembered for reflection or refraction at single
spherical surfaces, and for lenses of any kind or for any com
bination of them. The essential thing is to be consistent
during any calculation ; any inconsistency may lead to totally
erroneous results.
In all optical problems it is most important to remember
22 GEOMETEICAL OPTICS
that the symbol PC for a line, such as that in Figs. 9 and 10,
denotes the distance from P to C, so that it not only
expresses a length, but also the direction in which that
length is measured. Euclid uses the term PC as identical
with CP. We, however, must regard PC as a vectorial
symbol, and therefore equal to CP. Consequently
PC = PA + AC = PA  CA = AC  AP
Similar vectorial expressions will be used throughout this
book.
We have adopted the usual conventions that directions
from left to right are considered positive, and those from
right to left negative. Further, we shall regard directions
from below upwards as positive, and those from above down
wards as negative. As regards angles, we shall designate
the directions of rotation in the usual way. All rotations in
the counterclockwise direction are considered positive, while
all clockwise rotations are considered negative. The term
" Standard Notation " will in this book be used to denote
this device of signs to indicate these various directions.
In optics more blunders are due to the neglect of the
meaning of signs than to any other cause, so it is well worth
while devoting some attention to the subject. When a cor
rect mathematical formula is given, one knows that it must
be universally true, whatever values and whatever signs are
given in a special case to the algebraic symbols in the
formula.
Convex Spherical Mirrors. When the mirror is convex,
the point C, the centre, is behind the mirror, so that CA is
negative. A mathematician would know without any further
112
proof that the formula  f  =  must be true if the appro
p q r
priate sign were given to r. As, however, all our readers
are not necessarily mathematicians, we will give a formal
proof of this case.
As before, let C (Fig. 10) be the centre of curvature of
the convex mirror AK, and let the incident ray PK be
CONVEX SPHERICAL MIRRORS 23
reflected at K in the direction KR, so that the angle of inci
dence = $'. Produce RK to Q, cutting the axis in Q.
Then PKQ is a triangle, of which the exterior angle PKR
is bisected by the line CK, that meets the base produced
in C.
. PK_PC_PA + AC_PACA_^r
QK " CQ ~ CA + AQ " CA  QA ~ r  q (
When, therefore, a thin centric pencil is under considera
FIQ. 10.
tion, and PK and QK may be regarded as equal to PA and
QA, or p and q, we have, as before
P = P^L, which reduces to i + i = ?
q r q p q r
When the incident rays are parallel, i.e. when p = cc ,
1 121
 = 0, consequently  =  = ?, so they are reflected as if
P 2 T f
they came from a point F behind the mirror, such that FA
= JCA ; in other words, the principal focus is virtual and /
1121
is negative. We see, then, that the old formula + =  = 
still holds good if we assign the proper negative value to
r and /.
24 GEOMETEICAL OPTICS
Ex. An object is placed 8 ins. in front of a spherical
mirror, and an image of it is formed 4*8 ins. behind the
mirror. What is the radius of curvature of the mirror, and
what is its focal length ?
In this case p = 8 ins. and q = 4'8 ins., for the image
is behind the mirror. So we get, on substituting these
numerical values for the symbols, and paying due regard to
the signs they bear
1121 11
/./= 12 andr= 24
The negative signs show that both the radius and the
focal distance are to be measured in the negative direction,
i.e. both C and F are behind the mirror, that is the mirror is
convex.
This example will show the wealth of information that is
contained in this simple formula.
Geometrical Construction of the Image. Fig. 11 represents
a concave mirror, the centre of which is at C, and an object, AB,
on the principal axis of the mirror ; we are now going ibo show
how the image ab can be constructed, assuming, of course,
that only centric pencils contribute to its formation. As has
already been pointed out, the previous formula is only true
for thin centric or axial pencils, consequently in this case we
shall draw a tangent plane HOH' to the vertex of the
mirror, which will be a better representation of this centric
portion than the whole curved surface of the mirror would
be. In the figure, the line ACa may be taken to represent
a thin centric pencil as it passes through the centre C, but it
lies on a secondary, not on the principal, axis, and for this
reason the term " centric " is less likely to be misunderstood
than " axial." The plane HOH' will in future be called the
Principal Plane. We will give two methods by which the
image a can be found of a point A that is not on the principal
axis.
(A) Point not on the Principal Axis.
CONSTKUCTION OF THE IMAGE
25
(1) Draw ACa through the centre C, and draw AH to
the principal plane parallel to the principal axis ;
draw HFa through the focus to meet the line ACa
in a.
Then a, the point of intersection of HFa and ACa,
marks the position of the image of the point A.
(2) (Dotted lines.) Draw ACa through C, and draw
AFH' through the focus to the principal plane;
draw H'a parallel to the principal axis until it
meets ACa in a. The point a is the image of the
point A.
FIG. 11.
(1) The reason of this procedure will be apparent from the
following considerations. We know that any ray parallel to
the principal axis will be reflected through F, the principal
focus ; consequently, if AH represent such a ray it must be
reflected as HFa. Further, we know that any ray drawn
through the centre C must be reflected back along the same
course through C. Hence the point of intersection a must
represent the image of A. It is true that the line AH does
not represent any ray that is actually incident on the
mirror, for the point of its incidence would be so eccentric
that it would not be reflected through F. For this reason
the eccentric part of the mirror (suggested by spaced lines)
26 GEOMETRICAL OPTICS
has been covered up. We are, however, justified in asserting
that a small centric pencil from A will come to its conjugate
focus at the point of intersection a of HF and AC.
(2) We also know that light from any luminous point at
F must be reflected back parallel to the principal axis;
consequently the ray FH' must be reflected back as H'a,
so that a, its point of intersection with AC, must be the
image of A.
A vertical plane at F perpendicular to the principal axis,
such as FD, is called the Focal Plane, and has the following
properties: Light from any point on this plane will after
reflection travel in rays parallel to that axis on which the
point lies. For instance, light from D will be reflected from
the mirror as a beam of parallel rays in the direction H"5,
which is parallel to DC, the axis on which D lies. Also all
pencils of parallel rays that are but slightly inclined to the
principal axis will after reflection intersect in some point on
the focal plane. This property enables us to determine the
position of the image of a point that lies on the principal
axis.
(B) Point on the Principal Axis. (Spaced and dotted
lines.)
Through B draw any ray BDH", cutting the focal plane
in D; join DC and draw H"& parallel to DC, cutting the
principal axis in b. Then the point I is the image of the
point B.
Precisely the same construction can be applied to the
case of a convex mirror, which the reader is recommended to
draw for himself. It will be noted that whenever the
object is vertical, the image also will be vertical, so that
in drawing the image of AB in practice, all one has to do
is to find the position of a by either of the methods given,
and then draw a vertical line from it to meet the principal
axis in &.
Size of the Image. From a consideration of Fig. 11 it
is easy to find an expression for the height of the image (i)
as compared with the height of the object (0).
SIZE OF THE IMAGE 27
(1) Noting that ba is equal to OH', we find by similar
triangles that
*___&_ OH; _FO_ FO /
o ~ BA ~ BA ~ FB "" FO  BO ~/ p
(2) And, seeing that BA is equal to OH, it follows that
i*!L lOL F&  FQ ~ &Q /  9
o " BA " OH ~ FO " FO /
The formulae which we have now proved,  4 = 1,
dealing with the position of the image, and these
.
o fP f
dealing with the size of the image, can hardly be overrated,
since almost exactly the same formulae will be found to be
true when we come to deal with refraction at spherical
surfaces, while the method of construction of the image and
the method of determining its size are merely a repetition
of what we have just done.
It should be noted that all erect images are virtual, while
all inverted images are real.
An example or two will show the value of these formulas.
Ex. (1) A concave mirror has a radius of curvature
of 10 ins. What is its focal length ? An object 4J ins.
in height is placed 50 ins. :in front of the mirror. What
is the height of the image, and where is it formed ?
M
Here r = 10, so/ = ~ or 5 ins.
And since +=!, / = l _ / = ^"/ Q = fP
p q q P P pf
(This is the best form to use in all cases, as a similar
form holds good for refraction at spherical surfaces.)
28 GEOMETKICAL OPTICS
The position of the image is then 5 ins. in front of the
mirror, since q is positive.
Again, since  = r )  ==  ^ =  =  
o f po 5 50 45 9
The image is therefore inverted (as the sign is negative)
and real, and it is the height of the object ; i.e.
It is immaterial which formula for the size of the image
we use. Let us try the other formula :
So the height of the image is ^(4J), or ^ in.
Of course, the formula about size refers also to width.
Our second example is slightly more difficult.
Ex. (2) An object 6 cm. in width is placed at a
distance of 9 cm. from a reflecting surface. A virtual image
2*4 mm. in width is formed of it. What is the radius of
curvature of the reflecting surface ?
Since  = ^ , if  ip = fo :.f(i  o) = ip
and since the image is virtual, i is positive.
Consequently
2/* = 2i P  2 x 2 ' 4 x 9Q  4 ' 8 x 9Q
' ' ~ i  o~ 2460 57'6
48 60
6'4 8
As the sign is negative, the surface must be convex.
It may be noted that this is the basis of the method by
which the radius of curvature of the cornea is determined in
a living subject. A special apparatus is used to measure the
size of the image reflected from the surface of the cornea, and
from this measurement the curvature is calculated as in this
example.
GEAPHIC METHOD
29
Graphic Method for Spherical Mirrors. There is a very
simple graphic method for finding the position and relative
size of the image, which we will now give. The formula
 f
= 1 is so similar in form to the wellknown equation
P
x
If
to a line in terms of its intercepts  +  = 1, that it at once
suggests the following device (Fig. 12).
Draw two axes PH and HQ at right angles to each other,
and consider H the origin. On PH mark off a distance F'H
equal to the focal length / of the mirror, and draw the line
FT also equal to/ at right angles to PH.
FIG. 12.
As with a concave mirror / is positive, F'H and F"F
must be measured in the positive directions, i.e. either from
left to right or from below upwards (standard notation).
Suppose that an object is placed 20 ins. from a concave
mirror of focus 4 ins., and we wish to determine the position,
size, and nature of the image. Mark off the point P on PH
so that PH = 20 ; join PF", cutting HQ in Q.
Then QH is the distance of the image from the mirror ;
as it is measured from below upwards QH is positive, and it
is found to be 5 ins. in length, so the image is situated 5 ins.
in front of the mirror.
Again, * =j ; but/  p = F'H  PH = F'P
i F'H , . , . ... . 4
*  = ^F7r> > which in this case is ^
o r r lo
As F'H and F'P are measured in opposite directions 
30 GEOMETEICAL OPTICS
must be negative, so the image is inverted and real, and its
height is ^ of that of the object.
With the convex mirror /is negative, let/= 4 ins., so
FH and F"F are drawn as shown in Fig. 13, in the negative
direction. Suppose the object to be 16 ins. from the mirror,
PH is then made 16, and QH is found to be  3'2 (negative
because measured downwards). Consequently the image is
situated 3*2 ins. behind the mirror.
A A i F'H 4 1
Andas _=_ = __=_
the image is virtual and erect, because  is positive (F'H and
Fia. 13.
FT being measured in the same direction), and its height is
J that of the object.
*Eccentric Pencils ; Spherical Aberration, In the preced
ing sections we have been considering the reflection of small
centric pencils only ; when the incident light forms a wide
cone the several reflected rays cross each other at different
points, as is indicated in the diagram (Fig. 14). Paying
attention first to the eccentric pencil that falls on the mirror
at PQ, the reflected rays cross at FI and meet the axis in a
line at F 2 nearer the mirror than I, the conjugate focus of S
for centric pencils. The figure, of course, only represents a
section of the whole mirror ; if we consider it to be rotated
through a small angle about the axis OCS, the point FI will
trace out a small arc (approximately a line) so that at
FI and F2 two small lines will be formed : these are called
the primary and the secondary focal lines.
It will be noticed, also, that all the reflected rays that are
incident on the mirror from S touch a certain caustic surface
SPHEKICAL ABEKRATION 31
which has a cusp at I. This caustic curve may be commonly
observed on the surface of the fluid in a teacup, being formed
by the light that is reflected from the inside of the cup. A
caustic curve is frequently defined as the locus of the primary
focal lines ; this only means that it is formed by the points
of intersection of successive reflected rays.
The term " spherical aberration " is applied to all the
phenomena depicted in Fig. 14. If the reflecting surface had
not been spherical, but had been part of an ellipsoid of
revolution of which S and I were foci, all the light from S
that fell on this elliptical surface would have been accurately
FIG. 14.
reflected to the conjugate focus I, and the surface would then
be an " aplanatic " (airXavi'ig, not wandering) surface for the
source S. Similarly, a paraboloid would be aplanatic for
parallel rays.
The " longitudinal aberration " (IF 2 ) of a thin eccentric
pencil is the distance between the point where it cuts the axis
and the focus for a thin centric pencil.
Fig. 15 gives another view of the reflected pencil which
is shown in section as the shaded area in Fig. 14. The
incident cone is supposed to be pyramidal, so that the reflect
ing portion of the mirror concerned is of the rectangular
shape PQKKi. The reflected pencil is of a very curious shape,
the upper and lower rays (e.g. QFi and PF X ) of the two sides
32
GEOMETEICAL OPTICS
intersect in the primary focal line at FI, while the rays QF 2
and EF a from the upper edge intersect at F 2 the bottom of
the secondary focal line, and PF 2 and EiF 2 from the lower edge
at the top of the secondary focal line. Pencils like this
which do not converge to a point but to two focal lines are
called "astigmatic pencils" (a priv., crrfyjua, a point). As
FIG. 15.
we shall find that precisely the same condition arises with
eccentric pencils when refracted, it is well to spend some
little time in forming a clear conception of the solid figure
represented between the two focal lines at FI and F 2 . In
crystallography this sort of double wedge is called a sphenoid.
Near F 2 its section would be oblong, its height much greater
than its width; at D the section would be square, while
nearer FI it would again be oblong, but with its width greater
than its height. Clearly if the incident pencil were circular
REFLECTION AT SPHERICAL SURFACES 33
in section, the outline at D would be a circle. D is then the
position of what is called " the circle of least confusion," and it
is the best representative of the focus of an astigmatic pencil.
We will conclude this chapter by giving the formulae
without proof by which the position of the primary and
secondary focal lines can be determined of an incident thin
oblique or eccentric pencil on a spherical surface (Appendix,
p. 109). Let SP be denoted by u, FI? by v l} and F 2 P by v 2 ,
and the angle of incidence by <.
Then Ul.JL and l  + L =
u V rcos< u v r
QUESTIONS.
(1) An object 9 cm. in height is placed 10 cm. in front of a concave
mirror of focal length 25 cm. What is the height and character of the
image, and where is it formed ?
(2) The radius of a concave mirror is 16 ins. What is the distance
of the image from the mirror when the object is placed at a distance of
12 ins., and when placed at a distance of 4 ins. ?
(3) A virtual image onefourth the height of the object is formed
by a mirror. If the distance of the object be 9 ins. what is the radius
of the mirror ?
(4) An inverted image of a candle is thrown on a screen at a
distance of 6 feet from a mirror of focal length 6 ins. Where is the
candle placed, and what is the relative size of the image ?
CHAPTEE IV
REFRACTION AT PLANE SURFACES
So far we have been considering the path taken by light as it
travels through a single homogeneous medium ; we must now
find out what happens when light passes through more than
one medium. For the sake of simplicity, we shall limit our
selves to the consideration of homogeneous light, i.e. light of
the same wave frequency or colour, traversing different homo
geneous media. When light passes from one medium into
another, e.g. from air into glass, its course is altered, and the
light is said to be refracted. The subject was experimentally
investigated in 1621 by Willebroard Snell, who found that
there are two laws governing refraction :
LAW 1. The refracted ray lies in the plane of incidence.
LAW 2. The sines of the angles of incidence and refrac
tion are in a constant ratio for the same two
media.
The first law requires no explanation after what we
have already said about the similar law of reflection
( P . ii).
Students of physics will know that light in vacuo travels
at a rate of more than 186,000 miles per second, and that in
dense media its velocity is less. Indeed, in Optics the ex
pression dense medium merely means a medium in which
light travels with a relatively slow velocity. For instance,
water is dense as compared with air, but rare as compared
with glass , the speed of light in water is about f of its speed
in air, but about  of its speed in glass. Now, it has been
mathematically proved and experimentally demonstrated
that the constant ratio in Snell's law is identical with this
34
KEFKACTION AT PLANE SUKFACES
35
velocity ratio. If we denote the angle of incidence by <j> and
that of refraction by #', when light passes from water into
air its course is so altered that
sin
, = , i.e. about f .
v
When light passes from air into glass,
 approx
mately, for the speed of light in air is about f times its speed
in glass.
Let BB' (Fig. 16); represent the vertical plane that
forms the boundary of a dense
medium, glass for instance, and
let SC be an incident ray, CQ the
corresponding refracted ray, and
let NCM denote the normal at
the point of incidence C. The
original direction of the incident
light is indicated by the dotted
line OS', so MCS' is the angle
of incidence (ft, and MCQ is the
angle of refraction ^'. Note that
the angles of incidence and refrac
tion are to be measured from the
FIG. 16.
same normal, in this case CM. In the figure
S'L
and
sinf ~QM'
they are both positive, as of course they should be, for the
ratio  ^7 is the ratio of the speed of light in the first
medium to its speed in the second medium. As light is
reversible, we see that an incident ray in the direction QC in
the dense medium will be refracted at the surface BB' of the
rare medium as CS. Indeed, it follows from Law (2) that
all light on entering a dense medium is refracted towards the
normal, and on entering a rare medium away from the
normal.
The ratio of the velocity of light in a vacuum to its
velocity in a certain medium is termed the absolute index of
refraction of that medium, and it is denoted by the symbol
fi. Thus, if the absolute index of water be f and that of a
36
GEOMETRICAL OPTICS
certain kind of glass be f , it is easy to determine the relative
y w
index of water and glass, or .
339
= X T = x or
sm
4 8
Snell's law may be then stated as T* y , = 2 , where the
sin 0' /n'
subscripts denote the first and second media ; when the first
medium is air,  2. ^ 2 . The refractive index of air
differs so slightly from unity (being less than 1*0003) that
we shall commonly regard it as unity and denote it by JUQ.
Ee turning again to Fig. 16, the angle S'CQ is called the
deviation (D) of the light in this medium when the angle of
incidence or is MCS'. As SCS' is the original course of the
light, the deviation S'CQ or D is equal to  (MCS'  MCQ)
or D = (0 $') . This is clear, for the new direction <j>'
must be equal to (the old direction) + D. Annexed is a table
in which the angles of refraction and deviation are given
that correspond with certain angles of incidence at the sur
face of a kind of glass whose index of refraction ju is 1*52,
and again at the surface of water when /* = T333.
fA=l'52.
p = 1333.
$
<t>'
D
I
f
D
20
13 0'
 7 0'
20
14 52'
 5 8'
40
25 1'
 14 59'
40
23 50'
 11 10'
60
34 44'
 25 16'
60
40 31'
 19 29'
80
40 23'
 39 37'
80
47 38'
 32 22'
90
41 8'
 48 52'
90
48 36'
 41 24'
It will be noticed that when the angle of incidence in
creases uniformly, as in the first four rows, the angle of
refraction increases slower and slower; consequently, the
deviation ($' 0) increases faster and faster. It follows
from what has been said before that if the light be considered
as passing from the dense medium to the rare medium, we
CRITICAL ANGLE
37
must consider the <' in the above table as the angle of inci
dence, and $ the angle of refraction.
What will happen when the angle of incidence in water,
for instance, is greater than 48 36'? When $ is nearly
48 36' (Fig. 17) the refracted light just skims along the sur
face of the water. If be more than 48 36' the light will
be unable to leave the water, and it will be totally reflected.
The angle of incidence (48 36') at which this phenomenon
of total reflection occurs is called the critical angle.
It should be noted that when the path of light is from a
Fio. 17.
dense to a rare medium the relative index of refraction is
less than unity, for d ju r = ; in the case where yu r = 1,
d fjL r = . The critical angle is easily found for any medium ;
we have merely to give $' its maximum value of 90, and our
formula tells us that
I; thus sin
= df* r = , " the critical angle
= 48 36'.
Total reflection is frequently taken advantage of in the
construction of optical instruments, e.g. the camera lucida
(Appendix, p./^f ) and prism binoculars. Similarly when
the surface of water is viewed in a glass held above the
level of the eye, the silvery brilliance of the surface is due
38
GEOMETRICAL OPTICS
to total reflection. In aquariums the surface of the water is
often seen to act as a brilliant mirror.
We see, then, that part at any rate of the light in a rare
medium can always enter an adjoining dense medium, but
that when light in a dense medium is incident at any angle
greater than the critical angle, none of it will leave the dense
medium, as it will be totally reflected at the boundary of the
rare medium.
Image by Refraction at a Plane Surface. Let P (Fig. 18)
be a source of light in a dense medium (//) that is separated
FIG. 18.
by a plane surface AB from a rare medium (JUQ), say air.
Draw PAE normal to the surface. We will consider how the
point P will appear to an eye at E.
Since P is sending out light in all directions, we may
regard PA and PN as two of them. The ray PA will
undergo no alteration, for since PA is normal, both and 0'
must be 0. Draw the normal HNK at the point of
incidence N; then HNP = 0, and if HNQ be so drawn
that sm JE^ = ?, HNQ must be <i>'. Under these cir
sin HNQ IUL
cum stances produce QN to E, then the ray PN will be
refracted as NR. Now, if the figure be rotated round EP as
IMAGE BY KEFKACTION 39
axis, PN will trace out the limits of the axial incident cone
and QNK the limits of the emergent cone. Since HNP and
HNQ are equal to the alternate angles QPN and AQN,
o 1 _ sinHNP _ sin QPN _ sin QPN _ QN
fi' ' r fj! ~ sin HNQ ~~ sin AQN " sin NQP PN
As, however, we are only considering the narrow pencil that
enters the pupil of the eye at E, the point N will be very
close to A, and QN and PN may be regarded as equal to QA
and PA ; consequently, to an eye at E the position of the
"P A
virtual image of P will be at Q when QA = ^?PA = .
If, for example, a small object in water is viewed from a
point immediately above it, its apparent depth will be f its
real depth.
As in the case of reflection at spherical mirrors, this
determination of the situation of the virtual image is only
true when small pencils that are nearly normal to the
surface are considered. The formulae for oblique pencils are
more complicated, as two focal lines are formed with a circle
of least confusion between them (see Appendix, p. 111).
Refraction through a Plate. Let us consider how an
object P in air (JJLQ) will appear to an eye E in air when
viewed through a plate of glass (JUL) of which the thickness t
is AB (Fig. 19). The incident ray PN will be refracted as
NM on entering the glass, and on again entering the air at
M its course will assume the direction MR. As we are only
considering what will be the position of the virtual image Q
to an eye at E, we can find it very easily by the method of
the last section.
At the surface AN the point P will form a virtual image
i
at Q' such that Q'A = PA. But now Q' may be regarded
l*o
as the object for the second refraction at BM, and
Q'B = Q'A f t, consequently we have for the distance QB
from the distal surface
QB = Q'B = (Q'A + = PA +
40 GEOMETRICAL OPTICS
and since QA = QB  t,
When /MO =1 and // is denoted by ju, QA = PA 
\
and QB = PA 4 . Thus if fi = , QA = PA  1 or the
fj. A 6
image is formed nearer the plate by ^ of its thickness.
FIG. 19.
It will be noticed that the emergent ray MR is parallel
to the incident ray PN ; this is invariably true however
many media there may be, provided that they are bounded
by parallel planes, and provided that the final medium has
the same refractive index as the initial medium.
Succession of Plates. It is easy to see that, if any number
of media bounded by parallel planes are in succession, light
on emerging will pursue a course parallel to its original path
if the initial and final media have the same refractive
index pi. Let jui, ju 2 , ju 3 . . . denote the indices of the first,
second, third . . . medium respectively, and let 0i, fa, 3 . . .
denote the angles of incidence in these media ; then
sn
sn
sin
SUCCESSION OF PLATES
41
for, since the media are bounded by parallel planes, the angle
of refraction at one surface is equal to the angle of incidence
at the next. Thus, if there are four media interposed (Fig. 20),
and the angle of emergence into the final medium be denoted
.*. sin = sin b
or, in other words, the final and initial rays are parallel. In
FIG. 20.
the diagram the third medium is represented as less dense
than either of those adjoining it. Consequently, fa is greater
than ^>3 and ^g.
Prisms. Any refracting medium bounded by two plane
surfaces which are inclined at an angle to each other is
called a prism. The inclination of the faces BA and CA
(Fig. 21) is called the refracting angle or the apical angle of
the prism, and is usually denoted by A. The median vertical
plane that bisects the apical angle is called the principal
plane, while the plane CB at right angles to the principal
plane is called the base of the prism. We proceed to
42 GEOMETEICAL OPTICS
demonstrate certain properties which are common to all
prisms.
I. When light passes through a prism which is denser
than the surrounding medium, it is always deviated towards
the base.
If the face BM in Fig. 19 were rotated clockwise through
a small angle about B, the plate would form a prism with its
edge upwards and its base downwards ; the angles of incidence
and of refraction at BM will therefore diminish, and conse
quently any incident ray NM will on emergence be deviated
away from ME towards the base. If the rotation of BM be
continued, the angle of incidence at M, after passing through
the value of 0, will change its sign and become positive,
when a still more marked deviation of the emergent ray will
occur. Were the face BM rotated counterclockwise, the
base of the prism then formed would be upwards, and at the
same time the negative angle of incidence would increase,
and hence also the angle of refraction. The deviation of
the emergent ray would in such a case be upwards, towards
the base.
If the prism be less dense than the surrounding medium
the deviation is towards the edge of the prism.
II. As the angle of a prism increases the deviation also
increases.
This follows immediately from the proof given of I.
III. The apical angle is equal to the difference between the
angles of refraction and incidence within the prism.
When the apical angle A is measured in the same
direction as the angle of deviation D,
A = <//  $'.
In all the books on Optics A is said to be equal to </>' + ^', and in
order to get this result a new and special convention is adopted for the
signs attributed to <>' and \J/, which is never elsewhere employed. Our
convention about clockwise and counterclockwise rotation when
consistently used, will be found to give results that always hold good.
Let SIET denote a ray of light passing through the prism
BAG (Fig. 21). Draw the normals at I and E, and let the
PRISMS
43
angles of incidence and refraction at I be and <}>', and the
angles of incidence and emergence at R be $' and ^. Note
that in the figure <j> and $' are measured clockwise and are
therefore negative, while $' and $ are measured counter
clockwise and are consequently positive.
In the triangle AIR the angle IRA = 90  <//, and the
angle AIR = 90 + $', for $' is measured clockwise. Now,
FIG. 21.
B
since the sum of the interior angles of the triangle AIR is
180,
RAI + 90  f + 90 + 0' = 180
.'. A or RAI = $'  '.
and
carry,
This is universally true whatever signs
whether they be different or both the same.
IV. The total deviation is equal to the difference between
the angles of emergence and incidence less the apical angle of the
prism. D = ty A.
If the rays SI and TR be produced to intersect at D, the
exterior angle D of the triangle DRI is equal to the two
interior and opposite angles IRD and DIR.
44 GEOMETRICAL OPTICS
But IRD = i/.  V, and DIE or BID = (<j>  0')
.*. D = t   (i//  0') = ^  <  A.
It is clear that the total deviation D is the deviation of
the first surface, (0 0'), added to the deviation of the
second surface, <A $'.
As SIRT is the path of the light, the angle D is positive
(counterclockwise), and A is measured in the positive direc
tion also (RAI). If the light were travelling in the direction
TRIS the deviation D would be negative, and then A (or
IAR) would be measured in the clockwise direction. As
i and ^ would then be interchanged, it would be found in
that case also that
D = ^   A.
V. When a ray passes symmetrically through a prism, the
deviation is a minimum.
A ray passes symmetrically through a prism when the
angle of incidence is numerically equal to the angle of
emergence irrespective of sign, i.e. when < =  $.
If <f> increases, 0' increases also ; at the same time, how
ever, \j/' diminishes, and consequently ^ also. But as the
deviation (0  0') increases faster than the deviation
^ \j/' diminishes (p. 36), the total deviation must increase.
If we consider the path of light reversed, it appears that
when the angle of incidence is diminished the total deviation
increases. Hence the symmetrical position is the position of
minimum deviation.
When the prism is in the position of minimum deviation,
^ =  and $' =  <(>' ;
.. D = 2A  A, and A = 2^'
sin sin ^ sin i(A f D)
^ ij QJ r ' L i.
sin $' sin ^' " sin ^ A
This is the method used for determining the relative
refractive index of any transparent substance. The angles
D and A can be measured with great accuracy by means of a
PRISMS 45
spectrometer, and from these data the value of ju can be found
by this formula.
The prisms that are prescribed for spectacles are usually
very weak, i.e. the angle A is rarely more than 5, and for
these a simple approximate formula can be given for the
amount of deviation they induce. When $ and ^' are both
very small, sin $ and sin // may be replaced by ^ and if/,
and the formula for the position of minimum deviation
D = 2^  A may be replaced by 2/j.\j/' A or (/* 1)A, since
A = 2\j/'. This is a most useful formula if only used for
weak prisms when placed in the " symmetrical " position ;
thus, as the value of jj. for glass is little more than 1*5, we
find that the deviation produced by such a weak prism is
about half its apical angle.
Image by Refraction through a Prism. We have so far
been only considering the course of light refracted by a
prism ; when we try to form some idea of the appearance of
an object when viewed through a prism, we meet with several
difficulties that will only be briefly mentioned here.
We have seen that the deviation is not proportional to
the angle of incidence, and as an object of appreciable size
presents many points from each of which light is falling on
the prism at a different angle, it will be seen that the matter
is a very complicated one.
In Fig. 22 an object Pp is supposed to be viewed by an
eye in the neighbourhood of ET, the narrow pencil from P
that enters the eye being represented in the position of
minimum deviation. Clearly a virtual image of P will be
formed at Q, where the prolongations of the refracted rays
intersect. It is obvious that the image will be displaced
towards the edge of the prism, and that it will seem nearer
by one third the thickness of the prism traversed (p. 40).
Now, if we consider light emerging from p, it is clear that if
it fell on the prism in the position of minimum deviation it
would form a virtual image in the neighbourhood of q. But
an eye at E could not receive this pencil ; the only light that
would enter it must have had some such initial direction as
px, and the final image of p would be indistinct, for the
46 GEOMETRICAL OPTICS
lowest ray of the pencil entering the eye from p will have
undergone a smaller deviation than the uppermost one,
and consequently a confused image of p will
about q'.
FIG. 22.
It will be convenient to give a summary of the appear
ances of a square object when viewed through a prism in
different positions.
(1) When the plane of the prism is parallel to the plane
of the object, the edge of the prism being upwards.
The image is raised above the level of the object, the
sides being more raised than the midpart, so that the upper
and lower edges appear concave upwards.
(2) When the prism is rotated about a horizontal axis
parallel to its edge.
The image rises, and the height of the image is either
diminished or increased according as the edge of the prism is
turned towards or away from the observer.
(3) When the prism is rotated about a vertical axis, its
right side being turned away from the observer.
The right margin of the image is raised above its left
margin, so that the right superior and the left inferior angles
ROTATION OF PRISMS
47
FIG. 23.
are rendered more acute. If the rotation be in the opposite
direction, opposite results occur.
(4) When the prism is rotated about the sagittal line, i.e.
about the visual line of the observer.
The image rotates in the same direction, for, being always
displaced towards the edge, it follows it in its rotation. Let
a point of light be supposed to
fall on the centre of the cir
cular screen depicted in Fig.
23. When a prism that causes
a deviation R is interposed
with its edge to the right, the
point of light is deflected to
the periphery of the circle on
the right. If the prism be ro
tated through an angle /o, the
spot of light will move along
the arc through the angle p.
Consequently, the vertical displacement V will be R sin p,
and the horizontal displacement H will be R cos p. A
prism is frequently ordered by oculists in this way to correct
a vertical and a horizontal deviation of the eye simulta
neously. Clearly, if V and H are known the required prism
can be found, for R = \/V 2 + H 2 . Practically, it is found
what horizontal and vertical angular deviations are required
by the patient. Say that they are 9 and 0, then tan 8 = H
and tan $ = V, and a prism of deviation D is ordered such
that tan D = R, and set at such an angle p that
V tan
tan p = = 
H tan
In ophthalmic practice the prisms are so weak that we
may replace the tangents of 0, 0, and D by the angles (of
minimum deviation) themselves.
The reader is urged to verify the statements made in this
section by experiments with a prism, but a full explanation
of his observations can only be obtained in more advanced
treatises.
48 GEOMETRICAL OPTICS
QUESTIONS.
(1) Show why a stick that is partly in and partly out of water
appears bent at the surface of the water when viewed obliquely.
(2) It is found that when a plate of glass 7*7 mm. thick is placed
over a microscopic object the microscope must be raised 2  7 mm. to
bring the object into focus again. What is the refractive index of the
glass?
(3) If the apical angle of a prism be 60, and the minimum devia
tion for a certain kind of light 30, what is the refractive index of the
material of the prism for this light ?
(4) A prism of small apical angle (2), with refractive index T5, is
placed in water of refractive index ^. Show that its deviation is only
about onefourth of what it is in air.
(5) When viewing a distant object, each eye of a patient is found
to deviate outwards 1 44' (nearly V^) while the right eye deviates
above the level of the fixation line of the left eye 1. What two
prisms would entirely relieve this defect ?
CHAPTEE V
REFRACTION AT A SPHERICAL SURFACE
WHEN dealing with reflection at a spherical surface (Fig. 9,
p. 19), we considered the object P to be on the positive side
of the mirror. It would clearly make no difference to the
distance of the image from the mirror if the object were to
the right instead of the left of the mirror, except that the
image Q would then be situated to the right also. Since
optical formulae are universally true whatever values are
given to p, q, and r, when using the formulas it will generally
be found convenient to regard the direction of the incident
light as the positive direction. Hence PA or p may be
regarded as positive whether measured from left to right or
from right to left, and QA or q will be positive when both
Q and P are on the same side of the surface of the medium,
but q will be negative when Q and P are on opposite sides
of the surface.
When finding any general formula, to avoid error, it is the
simplest plan to use the " Standard Notation." As will be seen
in the following two sections, there is no difficulty in obtaining a
correct general formula when the object lies to the right of the
refracting medium.
We shall in Fig. 24 (and sometimes in future) consider
the object P to be situated to the right of the spherical arc
AK to familiarize the reader with the fact that the direction
in which light is travelling is quite immaterial when dealing
with such questions as these.
Concave Surface of Dense Medium. Let P be an object in
49 E
50
GEOMETRICAL OPTICS
a medium whose refractive index is JUQ distant PA or p from
a concave spherical arc, AK, bounding a medium of index rf,
and let CA or CK be the radius (r) of the arc ; then, if CKP
FIG. 24.
be the angle of incidence ft, and CKQ be the angle of
refraction ft',
ft __ sin ft _ sin CKP
fiQ ~~ sin ft' ~~ sin CKQ
Now, in the triangle PKC
PC  sin CKP _ sin ft
CK~sinKPC~sinKPC
and in the triangle CKQ
QC!
CK
sin CKQ
sin KQC
sm ft
smKQO
[Note that as PC, CK, and QC are all measured in the
same direction, the angles CKP, CKQ, and KQ C must all be
measured in the same direction, whether positive or negative.
In this case PC and CK are negative, so their ratio is equal
to the ratio of the sines of two positive angles.]
On dividing the first expression by the second, we get
PC __ sin ft sin KQC
QC = sin^'sinrKPC" 1
PA  CA p  r
n ^ f\K*  
*" QA  OA 2  r '
^ PK
sin KPQ "~ /u ' QK
PK
REFRACTION AT A CONCAVE SURFACE 51
This is universally true when p = PA, q = QA, and
r = CA, and when /IQ is the refractive index of the medium
in which P is placed.
When a thin centric pencil is considered, K must be very
close to A, and in such circumstances the distances PK and
QK may be regarded as equal to PA and QA, i.e. to p and q.
We then have for a thin centric pencil
p  r _ fi p
q  r~ fjto' q
or fiopq  fnqr = p'pq  ppr
On dividing by pqr we have
or
If now the figure be supposed to rotate round the axis
PCA, the extreme ray PK will trace out the limits of the
incident cone from P, and it is clear that all the constituent
rays of this cone will after refraction proceed as if diverging
from the point Q ; in other words, Q is the virtual image of
P. Indeed, the image of a real object placed in any position
before the concave surface of a dense medium is always
virtual.
When the first medium is air, /UQ = 1, and we get the
ordinary formula of the books :
q p r
This, however, is neither its simplest nor its most easily
remembered form, but before proceeding further we will
show that this formula applies equally to convex surfaces.
A mathematician would know, from the method of proof, that
the result must be universally true however the signs of p,
q, and r were changed, but we feel that this statement will
not be convincing to all our readers.
52 GEOMETEICAL OPTICS
Convex Surface of Dense Medium. As before, let PA = p,
QA = q, CA = r (Fig. 25), and consider the incident ray PK
having NKP for its angle of incidence, or < ; if then NKQ be
the angle of refraction, or 0',
i sin sin NKP sin PKG
sin 0' ~~ sin NKQ "~ sin QKC
Now, in the triangle PKC,
^ 4.u 4.  i mm Q sin
and in the triangle QKC, =
[A warning must here be given about signs : note that PC
is measured in the opposite direction to CK, so the angle
FIG. 25.
PKC is measured in the opposite direction to KPC, one
being clockwise and the other being counterclockwise, and
a similar precaution is necessary about the angles QKC and
KQC. Want of attention to points like these may give an
entirely erroneous formula.]
On dividing the first expression by the second, we get
sin

QC sm^'sinKPC /uo'sin QPK
PA  CA p  r _ ju' PK
QACA ^7
Consequently, in the limiting case when K is very near
to A
REFRACTION AT A CONVEX SUKFACE 53
pr ^p p
qr f*> q
or
The formula for a convex surface is therefore identical
with that for a concave surface, but whereas the latter makes
incident rays more divergent (unless p < r), a convex
surface always makes them more convergent, and this
depends on the fact that in the case of the concave surface
CA or r is measured in the same direction as PA, and in the
case of the convex surface in the reverse direction.
It will be noted that in both Fig. 24 and in Fig. 25 the
image Q is virtual, and hence P and Q are not conjugate in
the sense defined on p. 20. It would be found, for instance,
that if in Fig. 24 the object were placed at Q the image
would be virtual and situated somewhere between Q and C.
As stated before, P and Q are only interchangeable or con
jugate when the image Q is real.
In Fig. 25 we expect that if PA were increased a little so
as to render the incident cone less divergent, the refracted
ray KE would be parallel to the axis PAG. Now, if KR
were parallel to the axis, QA would be infinite, so let us put
q = oo in our formula.
Then L_^ = ^_ or / =  r
co p r fjL /io
We see, therefore, that when the source of light is put at
a distance p the emergent rays are parallel. This position
is called the First Principal Focus, and is denoted by F,
while the distance FA is denoted by/' instead of p'. We
have then/' =
Now suppose that the incident rays are parallel, or that
p is infinite,
K ' uo = t^^ or </=/
q oo r
54 GEOMETEICAL OPTICS
The point at which incident parallel rays come to a focus
is called the Second Principal Focus and is represented by
F" in the figure, while the distance F"A is commonly
denoted by /" ; so, replacing q' by its technical symbol, we
M  Mo
It should be noted that /" has the same sign as r,
and indeed F" is seen to be on the same side of the
surface as C ; F', however, is on the opposite side, as is
obvious from the expression for /' being preceded by a
negative sign.
Now, since the effect of a concave surface is to render
incident rays more divergent (unless p < r) t it will be clear,
FIG. 26.
on considering refraction at such a surface, that in order that
rays may emerge parallel they must have been initially con
verging towards the point F' (Fig. 26). Consequently the
First Principal Focus F' in this case is virtual. Again, the
Second Principal Focus F" is also virtual, as incident parallel
rays (the spaced and dotted lines) after refraction diverge as
if they were proceeding from F". It will be noticed that
here also/" has the same sign as r, while/' has the opposite
sign.
Sinc e ^L ,= 7 ^, /"r=/'
p.  ^ M  MO
KEFKACTION AT A SPHERICAL SURFACE 55
and we see, when the vertex of the spherical surface is
denoted by A, that both in Fig. 25 and in Fig. 26
F'A  CA = AF' or F'A, or/"  r = /'.
Note that both F' and F" are real when the dense
medium presents a convex surface.
Again, it is clear that
=  for =
/HO fi fio HQ ILL  JUG
We can now express our formula in its simplest form,
for on dividing
V_W> = H'VQ by P'"" weget
q p r r
f f"
J  + ^ = 1
P <1
f f"
This form, + = 1, is analogous to the expression we
have already found for reflection at spherical surfaces ; then,
indeed, there was no difference between the positions of the
first and second principal foci, as they both coincided in one
f f"
point. It will be found that y  + = 1 is much the easiest
P <l
form to remember, especially as we shall find that exactly
the same formula holds good with lenses.
Ex. (1). The refractive condition of the human eye may
be very closely approximated by a convex spherical segment
of radius 5 mm. bounding a medium of index f . Where
will its focus be for incident parallel rays, and what will be
the position of its First Principal Focus ?
Here the first medium is air, so /uo = 1 and // = J,
/ = ,  == y f = 20 mm.
and /' = ?/" = (20) = 15 mm.
56 GEOMETEICAL OPTICS
So F' is situated 15 mm. in front of the cornea, and F" 20
mm. behind it.
If the focussing mechanism were not called into play,
where would the image be formed by such an eye of an
object 16'5 cm. distant ?
/',/"_! ./"_*/' Pf"
f x . . ui q j,i
P q q p pf
Here p = 165 mm.
165 x 20 330 00

Presuming the curvature of the cornea and the other con
ditions of the eye to be normal, we should infer that if it
could not see an object distinctly at a greater distance than
16*5 cm., it must be 2 mm. longer than normal. We have
just found that a human eye that can see very distant objects
like stars distinctly must have its retina 20 mm. behind the
cornea.
Such problems are delightfully simple and easy, but mis
takes are frequently made when the object is in the dense
medium. By attending to the rule that HQ is used to indi
cate the index of that medium in which the object is situ
ated, all difficulty will be avoided. We will give an example
of the way such questions should be treated.
Ex. (2). There is a speck within a spherical glass
ball (r = 4 cm.) distant 2 cm. from one surface. Where
will its image be formed as seen from either side when
fi = 15 ?
The first point to remember is that the object will form
an image, whether real or virtual, irrespective of the presence
or absence of an eye to see it. Consequently, neglect the
position of the observer, except in so far as it determines
which surface of the sphere is being considered, and use the
" Standard Notation " for signs.
In this case the object P is in the dense medium, so JU Q
= 1'5, and the final medium is air, so p' = 1. Suppose the
ball to be so placed that the surface nearest to P is to the
left ; call this side A, and the distal side B.
REFRACTION AT A SPHERICAL SURFACE 57
A. Then PA or p = 2, and since TI = 4,
and
The image is virtual, within the glass, 1*6 cm. from A.
B. Here p' or PB = 6, and r a = 4 ;
so now/ 2 f or TL = 12, while / 2 " = 8
The image is virtual, and is formed at the distance of the
diameter from B, that is at the surface A.
In this case, as the second medium is to be regarded as
the refracting medium, it is obvious that at the surface B the
air presents a concave surface, and the incident divergent
cone from P is rendered less divergent, for QB is greater
than PB. Indeed, in every case (except when p < r, as in
case A) the effect of refraction at the concave surface of a
rare medium is to increase the convergence or diminish the
divergence. This can easily be remembered by considering
the case of a biconvex lens : convergence occurs at the first
surface, of course, because refraction occurs at the convex
surface of a dense medium ; and this convergence is increased
by the refraction at the second surface, which presents the
concave surface of a rare medium (air). When the rare
medium presents a convex surface, divergence results with
out any exception.
Geometrical Construction of the Image. The construction
is almost exactly the same as that we employed before, when
dealing with a spherical mirror. The only difference is due
58
GEOMETEICAL OPTICS
to the fact that now F' and F" no longer coincide, but are
situated on opposite sides of the refracting surface.
As before, we give two methods to find the image of a
point not on the principal axis, and one method for a point
that is on the principal axis. As we are assuming that only
centric pencils contribute to the formation of the image, we
FIG. 27.
draw the principal plane HOH" tangential to the spheri
cal surface at its vertex (Fig. 27). AB represents the
object, BOG the principal axis, F' and F" the first and
second foci of the spherical refracting surface whose centre
isC.
(A) Point not on the Principal Axis.
(1) Draw ACa through the centre C. Through A draw
a line parallelto the principal axis, meeting the
principal plane at H. Draw HF"& through F" to
meet the line ACa in a.
Then a, the point of intersection of HF" and AC
produced, marks the position of the image of the
point A.
(2) (Dotted lines.) Draw ACa as before through the
centre C; from A draw a line through the first
focus F' to meet the principal plane at H'. Draw
CONSTRUCTION AND SIZE OF IMAGE 59
H'a parallel to the principal axis until it meets
ACa in a.
The point a is the image of the point A.
It would appear, then, that the incident diverging cone
HAH' becomes, after refraction, the converging cone HaH'.
This is not strictly true, for the line AH does not represent
any ray that is actually incident on the refracting surface ;
but we are justified in asserting that a small centric pencil
from A will come to its conjugate focus at the point of
intersection of HF" and AC produced. Again, we know
that all rays that pass through F', such as AF'H', must,
after refraction, proceed in a direction (H'a) parallel to the
axis.
(B) Point on the Principal Axis. (Spaced and dotted
lines.)
Through B draw any line BDH" cutting the first focal
plane in D and the principal plane at H" ; join DC and draw
H"6 parallel to DC, cutting the principal axis in &.
Then the point b is the image of the point B.
The reason of this construction is obvious from the pro
perty of the focal planes described on p. 26. Light from
any point on this plane will, after refraction, travel in rays
parallel to that axis on which the point lies. Now, DC is,
of course, the axis on which D lies, and consequently the
incident light ray DH" must, after refraction, take the
direction H'7> parallel to DC.
Size of the Image. In just the same way as we found the
height of the image formed by reflection at a mirror, we can
find  in this case.
(1) Noting that la is equal to OH', we find, by similar
triangles, that
i = fa_ = OH' _ FO _ F'O /'
o ~~ BA "" ~BA ~~ FB "" F'O  BO "f^^p
(2) And as BA is equal to OH,
!=!?L JOL F"fr _F"050
o BA ~ OH " FO ~ F'O
60 GEOMETEICAL OPTICS
Exactly the same construction is used when the refracting
surface is concave, in which case F" with C would lie on the
object side of the refracting surface, and F' would lie on the
opposite side. It should be noted, also, that when the refract
ing medium is dense and presents a concave surface the
image is erect and virtual, but when the surface is convex
the image is inverted and real as long as p is greater than /' ;
if p =f the refracted rays are parallel, and if p </' they
are divergent. We will give one or two examples where JJLQ = 1
(the first medium), and fi = f .
Ex. (1). A refracting surface presents a convex surface of
radius 4 cm., and at a distance of 8 cm. in front of it is
placed an object 5 cm. high. Give the position, the character,
and the height of the image.
We must first find the foci
=M== =12 cm.
and /" =  /' = (12) =  16 cm.
The image is therefore formed 32 cm. on the object side of
the refracting surface, so it is virtual.
I 12
_
o f'p 128""
or if we use the second formula
i_f"q_ 16 32
~o~~f~ 16
The same result must be obtained in either case : the
image is erect because  is positive, and three times higher
than the object. Consequently, the height of the image is
15 cm.
This is a case in which p is less than/', and the refracted
GRAPHIC METHOD 61
rays diverge, though less than the original incident rays, so
a virtual image is formed at a greater distance than that
of the object.
When speaking of the effect of concave refracting surfaces,
we said that incident rays were always made more divergent
unless p were less than r. As an example, we will now
take a case where p is less than r.
(2) An object 5 cm. high is placed 3 cm. in front of a
concave refracting surface of radius 4 cm. Where will the
image be formed, and what will be its height ?
In this case, as the surface is concave r is positive ; so
The image is therefore situated 3*2 cm. in front of the
surface, or 2 mm. further off than the object ; in other words,
the divergent cone from each point of the object is made
rather less divergent,
and
So the image is virtual and erect, and as the object is
5 cm. high, the height of the image is 4 cm.
Graphic Method for Refraction at a Spherical Surface.
The graphic method that we employed for spherical mirrors
can also be very conveniently used in this case. As an example
we give in Fig. 28 the method as applied to the normal human
eye. The complex system of the eye with its cornea and lens
will be shown on p. 102 to be almost exactly equivalent, from
an optical standpoint, to a single refracting convex spherical
surface of radius 5'25 mm., with /' or F'H equal to
+ 1554 mm. and/" or F"H equal to 2079 mm.
On the horizontal straight line PH we mark off the
point F' so that F'H = 15*54 mm., and from F' we raise a
vertical line such that FT = 20*79 mm., for we consider
62
GEOMETKICAL OPTICS
lines measured from left to right and from below upwards
as positive, and those in the reverse directions as negative.
Where will the image be formed by such a refracting system
of an object 8 cm. distant ?
Make PH = 8J cm., and join PF" and produce it to
cut HQ in Q. Measure QH; it is found to be 2556 mm.
The image is therefore formed 25'56 mm. behind H, which
denotes the cornea. Now, the " standard " eye that can see
F"
FIG. 28.
very distant objects distinctly must have its retina at a
distance of 20*79 mm. from its cornea ; suppose that a cer
tain person could not see objects further off than 82*5 mm.
(this case), then his retina must be 25*56 mm. from his
cornea, or his eye must be 4*77 mm. too long, if his eye
were otherwise normal.
The size of the retinal image is given by our previous
f , i F'H 1554
formula,  = ^^ = ^757. , so that the image is inverted
and real, and a little less than onefourth of that of the
object.
Further, our diagram will tell us what glass would correct
this eye for distance. Spectacles should always be worn in
the first focal plane of the eye, i.e. about half an inch in front
of the cornea, so that F' marks the position of the correcting
GKAPHIC METHOD 63
glass, and it must be of such strength that parallel rays from
a distant object should form their image at P. This means
that the second focal distance (/") of the correcting glass for
distance must be equal to PF', which is found by measure
ment to be 67 mm. approximately. Now, a lens of which
1 000
/" = 67 mm. is one of power ^= y or nearly 15D, as
will be shown in the next chapter. Indeed, if preferred, this
division sum may be done graphically in this way. Draw a
line F'A 50 mm. long in any direction that makes an acute
angle with F'P (Fig. 28); mark off S on PF', making SF'
equal to 20 mm. Join AP and draw SD parallel to AP ;
then F'D in millimetres gives the power of the correcting
glass in dioptres (see p. 80). For of course
F'D _ F'A . F'D _  50 _ 1000  
SF' " PF" l ' e ' "20 = "67" ~67~
Note that both F'D and F'A are measured in the negative
direction.
If the eye were 3'23 mm. too short, where should the
object be to form a distinct image on the retina ?
Make QH 17'56 mm. (i.e. 2079 + 323). JoinF'Q,
and produce to meet the base line in P (Fig. 29). If accu
rately drawn, P will lie 84'46 mm. to the right of H. This
means that PH is negative ; thus, unless very great focus
ing power were used, the eye would not be able to see any
real object distinctly, for light would have to converge as if
towards a point 8446 mm. behind the eye in order to come
to a focus on the retina. It is unnecessary to determine the
size of the image, as we have seen that no real object could
be seen. The correcting glass must have /" equal to PF',
which by measurement we find to be  100 mm.
*=.+
Or if we wish to do this division graphically, we draw F'A
50 mm. long in any direction, making an acute angle with
64
GEOMETRICAL OPTICS
FP. We then mark off S between P and F', making SF
equal to 20 mm. Then by joining PA and drawing SD
parallel to it, cutting FA in D, we measure FT). We find
A
FIG.
that F'D = 10 mm., so his correcting glass is 410D. The
graphic method is given in this case to show the generality
of the method, though here there is no special advantage in
it. As before, we have
F'D F'A
F'D
50
PF" l ' e ' 20~ 
 J ~
1000
100
*Eccentric Pencils Focal Lines. In the preceding sec
tions we have been considering the refraction of thin centric
pencils, of which the central ray traversed the centre of the
refracting surface. We will now take the case of a thin
eccentric pencil that does not pass through the centre.
Let be a luminous point on the axis of the concave
surface whose centre is at C (Fig. 30). The thin pencil
POQ will be refracted in the direction RR', so that the
refracted rays, if produced backwards, will meet at FI, and
will cut the axis in a line in the neighbourhood of F 2 , so
that the refracted pencil will be astigmatic. If the incident
pencil were of a pyramidal shape, the prolongations of the
refracted rays would form a figure something like Fig. 15,
only in the case shown in Fig. 30 FiP is greater than F 2 P.
A similar sphenoid will be formed between the secondary
REFRACTION AT A SPHERICAL SURFACE 65
and primary focal lines, but the lateral surfaces of the
refracted beam will now intersect in the secondary focal line
at F 2 before the final edge of the sphenoid is formed at F x ,
the primary focal line. If OP were greater than ^(FxP),
F 2 P would be greater than FX? (see Appendix, p. 116); in
such a case the refracted beam would take exactly the form
represented in Fig. 15.
The formulae for the lengths FiP or Vi, and for F 2 P or v 2 ,
when the source of light is at a distance OP or u t will be
merely stated in this place
/u cos 2 ft' _ cos 2 ft __/*__!_ sin (ft  ft')
r sin ft'
FIG. 30.
The proof of these formulae will be found in the Appendix,
pp. 114116.
Conversion of Formulae of Refraction into those of Reflec
tion. On p. 35 it was stated that . , { or )= ^, i.e. the
sin ft \ HQ} V 2
ratio of the speed of light in the first medium to its speed in
the second medium. Now, when reflection occurs at a sur
face, it is clear that the speed of the light is unchanged in
absolute amount, for it is in the same medium, but its direc
tion is reversed ; in other words, the sign is changed, so that
7
66
GEOMETKICAL OPTICS
for reflection we must consider V 2 as equal to Vi. We
' "XT' ~\T
have therefore or ^ = ~ = 1, so that every formula
/"o V 2  V i
applying to refraction must apply to the similar case when
reflection is considered, if we simply replace the expression
i
for the relative index of refraction by 1. Indeed, this
/*o
is one of the best tests we have to determine whether a for
mula for refraction is correct or not. Eeplace // by  1 and
IHQ by 1, and see if the correct formula for reflection is given.
We will give a few examples to illustrate the generality of
this method of conversion.
Refraction.
Reflection.
sin <f> /i'
sin </>' ~JL^
sin _ 1
.,=,
sin ^'
/*' /*0 "' ~ Mo
1 l_2 ^1^1_2
gj) r
2 P r
1> 2 *
/* 1 sin (<f> $')
 1 1 sin 24
_ 2sin0cos<J>
V 2 u ~ r sin <f>'
v t u rain
11 2 cos
^> "~ r sin ^
v z + u r
/u cos 2 ^' cos 2 <f> sin (^ ^')
COS 2 COS 2 (ft
2 sin ^ cos
1). % 81H rf>
v l u
or 1 + 1
r sin ^
2
v,u~
r cos ^
The reader need not therefore burden his memory with
any of the special formulae for eccentric reflection, as the
conversion of the formulae for eccentric refraction is so easy.
As another aid to memory, I may add that any formula for a
curved surface becomes true for a plane surface if we make
r = oo , for this is equivalent to making the curvature 0.
For instance, taking our last illustrations for eccentric
pencils
fi cos 2 V _ cos 2 ft __ fji _ 1 _ sin(<ft $')
~
u
v<i
u
r sn
REFRACTION AT A SPHERICAL SURFACE 67
becomes, when r is made infinite
COS 2 ft _/f_ _ 1 _ Q
U V% U
COS 2 ft'
or #1 = UM a 1  and t^ = M^
cos^ft
These are the distances of the primary and secondary
focal lines when a thin oblique pencil undergoes refraction
at a plane surface, fj, denoting the relative index of refraction
(Appendix, pp. 111113).
QUESTIONS.
(1) Assuming that the human eye is a simple refracting system, of
which the first focal distance is 15 mm. and the second focal distance
20 mm., where would the image be formed, and what would be its
height, if an object 6 mm. high were at a distance of 150 mm. ?
(2) What curvature must be given to the refracting surface when
/* = J in order that the previous object at a distance of 150 mm. may
form a real image at a distance of 20 mm. ? What would be its
height ? Compare the size of the retinal images in the axial myopia
(the first case in which the eye is long enough to form a distinct
image of the object) and the curvature myopia (the second case).
(3) If the back of a glass sphere Q* = f ) be silvered, where will be
the image that is formed by one reflection and one refraction of a
speck that is halfway between the centre and the silvered side ?
CHAPTER VI.
LENSES.
ANY refracting medium bounded by two curved surfaces
which form arcs of spheres is known as a spherical lens;
the axis of the lens is the line joining the centres of the
spheres, and that part of the axis lying between the two
surfaces gives the thickness of the lens.
Thus in Fig. 31 the surface A is the arc of a sphere that
has its centre at C b and the surface B is the arc of another
FIG. 31.
sphere that has its centre at C 2 : the axis of the lens is the
line joining Ci and Cg, and the thickness of the lens is the
length of the line AB. These are the only definitions we
shall require for the present. It will be apparent that if one
of the surfaces be plane, it may be regarded as a spherical
surface of which the radius is infinite.
Thin Lenses. Thin Axial Pencils. The conjugate focal
distances of an axial pencil of a thin lens can be very
easily determined from the formula that we have already
found for a single spherical surface. It should be noted that
the term " centric " is no longer equivalent to the term
" axial," as an oblique pencil that passed through the centre
of one spherical surface would not pass through the centre of
68
LENSES 69
the other surface unless they coincided (see p. 84). We
shall eventually determine the position of what is called the
Optical Centre of a lens, and discuss its properties; at
present we are only concerned with axial pencils.
Let P (Fig. 31) be a luminous point on the axis of the
lens, and let Qi be the image of P due to refraction at the
r
first surface A, and let be the relative refractive index
Mi
between the first medium and the second medium (e.g. air
and glass). When p = PA, qi = QiA, and when r x is the
radius of the first surface,
qi p n
The light from P, after traversing the first surface A, will be
travelling in the direction towards Qi. On emerging at the
second surface into the original medium (JUQ), refraction again
takes place, and the relative index is now ^? . As the lens is
considered to be of negligible thickness, QiB may be regarded
as equal to QxA or q im
Eegarding, then, Q! as the source for the second refraction
at B, of which the radius is r 2 , and putting q for QB, the
distance of the final image, we have
i f
JUQ JH fJ.Q fj.
q qi r z
On adding (a) t_^P = ^Jfo
Mo fto , t
we get = (ju
This is the standard formula for a thin lens ; when the
lens is in air (as is almost always the case) /^ = 1, and we
can suppress the dash and write
1 1 T /l 1\
=!( ) . . . . (c)
q p \n r 2 J
To find the First Principal Focus (F r ), as before, we make
70 GEOMETEICAL OPTICS
the emergent rays parallel, or we make q infinite ; we then
have  = 0, and writing / ' for this special value^ of p } we
get
To find the Second Principal Focus (F"), or the focus for
incident parallel rays, we make p infinite ; so, substituting
/" for this special value of q, we get
/"
We see that /' = /", which is always the case when the
initial and the final media are the same ; so, on substituting
these values in (c), we find that
111 /" /"
= or ' *. = l
q p f" q p
and finally our old formula
+^=1 (A)
p q
[It may be here noted that when thick lenses are con
sidered, if the thickness be denoted by t, the term
must be added to the expressions for ry, and subtracted from
that for ^7 .
Thus
and
These expressions give the focal lengths correctly, and for
some purposes this is all that is required, but they do not
LfiNSES 71
enable us to determine the position of F and F". We shall
find later (p. 91) that when a thick lens is considered, there
are two Principal Points towards which the two focal dis
tances must be measured respectively.]
Let us suppose that the lens in Fig. 31 is of negligible
thickness and has for its radii of curvature r x = 2 cm. and
r a = 4 cm. What will be its focal distances when /u = 1*5 ?
1/1
/./' =f cm. and /" = f cm.
Consequently, if a source of light were placed at 2 cm.
distance from such a lens, the emergent rays would be
parallel. If, however, the incident rays were parallel, as for
instance from the sun, they would converge to a focus 2 cm.
on the far side of it, as the negative sign shows that the
Second Principal Focus is behind the lens.
Ex. (1) If an object were placed 24 cm. in front of such
a lens, where would the image be formed ?
Here p = 24, and we wish to find the corresponding
value of q.
f+r =l ,r = P^L or q= *r
p q q  p pf
' 24() 24 x 8
80 2= 2T^ = ~6r~ = m 
The image would therefore be real, and it would be formed
3 cm. behind the lens.
(2) An object is placed 24 cm. in front of a concave lens,
of which TI = 2 cm. and r a = 4 cm. (n = 1*5). Where
will its foci be situated, and where will the image of the
object be formed ?
This lens has the same curvature as that in (1), but in
the reversed direction, so/' will be found to be f cm. and
Hence q or = = 24 cm.
72 GEOMETKICAL OPTICS
The image in this case is virtual, and is formed 2*4 cm.
in front of the lens.
It should be noted that in all converging lenses /' is
positive (i.e. F' is on the same side of the lens as the object
P) and /" negative, and they both are real, whereas in all
diverging lenses /' is negative (i.e. F' is on the side opposite
to P) and / " positive, and they both are virtual. Moreover,
all converging lenses are thickest in the middle, whereas all
lenses which are thinnest in the middle are diverging in
function. The reverse of these statements is not true,
although it is frequently alleged to be so ; e.g. a lens can be
constructed that is thickest in the middle and yet be diverg
ing in function.
When a lens is converging in function, the image is real
and inverted (Fig. 32) or virtual and erect (Fig. 33), accord
ing as the distance of the object is greater or less than the
first focal distance of the lens. With diverging lenses the
image of a real object is always virtual.
In Figs. 32 and 33 the object is represented to the right
of the lens, and therefore F' must be on the same (the
incident) side, and as the lenses are of the same focal length
(/' = 4 cm.) we make OF' in each case equal to 4 cm. In
Fig. 32 the object AB is placed 14 cm. from the lens, so the
image ab will be real and inverted, and it will be situated at
a distance of 5'6 cm. on the other side of the lens,
for
.
p~f 144
In Fig. 33 AB is placed 3 cm. from the lens, and the
image ab is therefore virtual and erect, and it is 12 cm. from
the lens on the object side.
P / 34
Geometrical Construction and Size of the Image. The
image is drawn ' by the same method that was employed in
constructing the image formed by refraction at a single
spherical surface (Fig. 27, p. 58). Three alterations are,
CONSTRUCTION OF THE IMAGE
73
however, necessary : (1) the principal plane HOH' must be
drawn through the centre of the thin lens instead of tangen
tially to its surface ; (2) OF" must be made equal to FO ;
and (3) the point must be regarded as taking the place
of C, which is no longer required (cf. Figs. 32 and 33).
FIG. 32.
When the lens is symmetrical, as in Fig. 32 and 33,
O is the optical centre of the lens, and, clearly, any ray of
light that traverses proceeds on its course without devia
tion, for any refraction that it may undergo on encountering
the first surface of the lens will be reversed on emerging
from the second surface. It is clear from Fig. 32 that the
ray AOa cuts the lens at two points such that the tangents
at the points of entry and emergence are parallel, so that
the light traverses the lens as if it were a plate with parallel
sides.
(A) Point not on the Axis. The line AO is drawn and
produced to a. (1) AH is drawn parallel to the axis, meet
ing the principal plane in H ; HF" is then drawn through F"
meeting AOa in a (Fig. 32) ; or (2) F ; A is joined and pro
duced to meet the principal plane in H (Fig. 33), and from
H the line Ha is drawn parallel to the axis, meeting OAa
in a.
74
GEOMETRICAL OPTICS
(B) Point on the Axis.From B a line BDH' (Fig. 32) is
drawn at any acute angle cutting the first focal plane in D
and the principal plane in H' ; DO is then drawn, and H7> is
drawn parallel to DO.
From Fig. 32 we see that
F"Q  t>0 _f" ; q
F"0 /"
FIG. 33.
F'O
and from Fig. 33 that
i _ la _ OH _ F'O
o ~~ BA ~ BA " F'B ~ F'O  BO ""/'  p
Therefore when F'O or/' = 4, BO or^ = 14, and when BA
is 8 inches in height, as in Fig. 32,
f 4
i or la = ^T BA = T
fp 414
8= 32
But if BO or p = 3 and BA = 2, as in Fig. 33,
^2 = 8
f f"
The formulae J  + = 1
p q o j p j
thus shown to be universally true for refraction at a single
SIZE OF THE IMAGE 75
spherical surface, for lenses, and also for reflection, when it
is remembered that in the case of a mirror F' and F" coin
cide in one and the same point F.
Figs. 32 and 33 show, however, another expression for
 that is always true in the case of lenses, for ^r = ^^ = 
1>A JhJU p
It is, however, rarely necessary to pay any attention to this
relation, and it is much easier to remember the few formulae
that are universal, which are given above.
There is, however, one important point to which atten
tion may be particularly directed, as it affords the explana
tion of certain facts which are obvious enough if it be borne
in mind, but puzzling otherwise. It is this, that the angle
BOA subtended by the object at the centre of the lens is
always equal to the angle bOa subtended by the image. The
application of this fact may be illustrated by the following
example : the diameter of the sun subtends a visual angle of
31' : what is the diameter of its image as formed by a lens of
1 m. focal length ? (tan 31' = 0'009.)
In this case, of course, an inverted image is formed at
F", and
la or i = /" tan 31' =  1000 X 0'009 mm. =  9 mm.
In the case of a mirror, it may be noted that the angle
bOa is numerically equal to the angle BOA, as may be at
once seen by drawing the lines aO and AO in Fig. 11.
Then if bOa be denoted by a, BOA = a, for it is measured
in the reverse direction to bOa, so that i = q tan a. When
distant objects like the sun or the moon are observed with
a reflecting telescope, the incident rays are parallel, and the
size of the image formed by the mirror is given by the
expression
i = / tan a.
Ex. (1). A camera of 6 inches focus shows a distinct
image 1 inch high of an object when the groundglass screen
is 6*6 inches away from the lens. At what distance is the
object, and what is its height ?
76 GEOMETRICAL OPTICS
Here an inverted image is formed 6*6 inches behind the
lens, so
:.p or **~*s = g g , c = 66 ins., i.e. 5 feet 6 ins.
'
q = 6'6 ins.
**~*s = g g , c
q j o'b + o
A i /"  1 6 + 66 1
and o = V' ^T =6 = ~io
/. = 10 ins.
Ex. (2). The same camera shows an image 1 inch in
height of a man who is 6 feet high. How far off is he ?
i f 16
_ v m* _ __ _
o~fp 72 ~Gp
:.p  6 = 6 x 72 ins. or p = 36J feet.
Ex. (3). If in repairing a bicycle reflex lamp the plane
mirror is placed at the focus of the convex lens, will it act
in the desired way ? No. The light from a distant approach
ing motor that is incident upon the lens will converge to its
focus. From this point on the mirror the reflected light will
fall again on the lens, and will emerge as a parallel beam by
its previous path. It will therefore only return to the
source of light, and hence will give no warning to the driver
of the car.
It is for precisely the same reason that the pupil of the
eye appears black. It is really red when incident light
falls upon it ; but the observer necessarily puts his head in
the path of the incident light in one direction, so that no
light is returned in that direction.
In order to see the back of the eye, an ophthalmoscope is
used, which consists essentially of a small mirror with a
central perforation, through which the surgeon looks at the
eye of the patient. Light is reflected by this mirror into the
patient's pupil to the red background of the eye, and return
ing by its previous path to the mirror is received in part by
the surgeon's eye behind the aperture, so that the pupil
appears of a bright reddish colour to him. If the retina of
the patient be not situated at the focus of his eye, the light
MAGNIFICATION
77
on its return from the retina will no longer emerge as a
parallel beam, but either as a convergent or divergent pencil,
according as the retina is behind or in front of the focus.
In such cases, especially when the pupils are widely dilated,
they will appear to glow with a ruddy light, even to the
unaided eye. For a similar reason, the unsatisfactory reflex
lamp may be made efficient by moving the plane mirror a
little nearer the lens, so that it is just within its focal
distance.
It has been shown that in Fig. 33 the image ab is four
times the height of the object AB, but this does not neces
sarily mean that the apparent size of this virtual image, to
an eye that perceives it, is four times the apparent size
of AB.
Magnification. The apparent size of an object depends
upon its distance from the eye that perceives it. If we
regard the eye as a simple refracting medium formed of a
single spherical surface of radius 5*25 mm., as on p. 61,
the centre of this surface will represent the nodal point (K)
of the eye 5 '25 mm. behind the cornea. Now, an object BA
FIG. 34.
(Fig. 34), at a distance KB from the nodal point, will sub
tend at K the angle BKA or 9. This is called the visual
angle. The apparent size of BA is evidently determined by
T> A
tan or . It is obvious that tan 6 could be increased
indefinitely by diminishing KB indefinitely. In other words,
the apparent size of an object could be indefinitely increased
by bringing the eye indefinitely close to it. The eye, how
78 GEOMETRICAL OPTICS
ever, is incapable of seeing an object distinctly which lies
within a certain distance. The distance KB of this punctum
proximum, as it is called, from the eye varies in different
individuals, and increases with age, so that it is impossible
to assign to it any definite value which shall be applicable to
all cases. If I denote the least distance for the individual
eye considered, then the greatest value that tan 6 can actu
ally have is by making KB equal to I, when tan 9 = j or y
It is customary to assign an arbitrary value of 10 inches to I.
If, now, a convex glass be placed as in Fig. 33, with the
object BA within its first focal distance, and the nodal point
of the observer's eye be at K, a virtual image la will be
formed at a distance "Kb from the nodal point. Then if 0' is
the visual angle subtended by la at K, tan 6' = T^T or ^.
But we have already found that the maximum size of the
object as seen by an unaided eye is given by tan 6 or j
The magnification M of a convex glass must therefore be the
relation between tan 0' and tan 9.
_ tan 0' _ i I _ l_ /lJ^JL/'i L^\
~ tan ~~ K6 ' o ~ Kl ' /" ~ Kl\ f)
When a convex lens is used as a magnifying glass, the
image is always virtual, and it therefore lies on the same
side of the glass as F', so the fraction ~ is always positive
f' ~ FO = OF"')' ^ * S C * ear k m ^ e a ^ ove ex P ress i n
that M is increased by making KZ> as small as possible, and
by increasing the value of q. But K& cannot be made
smaller than I (say, 10 ins.), or the image would not be seen
distinctly, and as "Kl = KO + 01, say, d j 05, there is a
limit to the value of 01. The distance d of the lens from
the nodal point of the eye cannot be much less than half an
inch, so we must make 01 = 9 J inches in order to get the maxi
mum magnifying power out of a lens employed in this way.
MAGNIFICATION
In the case above, with a lens of 4 inches focus, with the
object so placed as to form a virtual image 9J inches from
the lens, the greatest amount of magnification will be
obtained, viz.
M =
There is another way in which a convex lens may be
used as a magnifying glass, viz. when the object is placed in
the first focal plane of the lens (Fig. 35). HOH' is the
FIG. 35.
principal plane of the lens, which it is unnecessary to indi
cate, and the object BA is placed in the first focal plane, so
that the incident cone of rays from B (H'BO) will, after
refraction, emerge as a beam parallel to the axis, while the
incident cone from A (OAH) will emerge as a beam parallel
to AO. If, now, an emmetropic eye (or an eye adapted to
see very distant objects) be situated behind the lens, an
image of BA will be formed on its retina. It is clear that
in this case the apparent size of BA will be independent of
the position of the eye, for its size depends on tan BOA or
tan 0". If, however, the eye be situated behind the point X,
it is obvious that it could not see the whole of the object, as
only rays from the central part of BA would enter the eye ;
in fact, on increasing the distance of the eye from the lens
the field of view would be diminished, but there would be no
alteration of the magnification of that part that was seen.
80 GEOMETEICAL OPTICS
As BA is in the first focal plane, tan 0" = j, and so the
tanfl" o I I
magnification M in this case is  s = ~ f . = JF .
tan / o /
Our lens of 4 inches focus, used in this way, would only
give a magnification of j, or ^ = 2*5, but it would be less
fatiguing to the eye.
On comparing the magnifying power of a convex lens
used in these two different ways, we see that in the first case
it is 1 f j n and in the second case it is j, or , ^. There
fore, if d be less than /' the first method gives the higher
magnification, and vice versd.
Graphic Method for Lenses. A method similar to that
illustrated in Figs. 28 and 29 may be employed for finding
the position and the relative size of the image formed by any
lens. Fig. 28, when F"F' is made equal to F'H, will represent
the construction necessary for a convex lens whose first focal
distance (/') is indicated by F'H, whereas Fig. 29 if inverted
would give the construction for a concave lens when F"F' is
made equal to F'H, its focal distance. Such methods will,
however, be seldom found to be of practical use, as we have
vf " i f
such simple formulae as a = * J ^ and  = ^ ready to
P~f o f  p
our hand for the solution of questions like these.
Power of a Lens. Dioptres. In a previous section we
used the expression " magnifying power," and we saw that
itivaried inversely as/' ; we have seen, too, that strong lenses
with strongly curved surfaces had short focal lengths. We
are therefore quite ready to admit that p is an adequate
j
measure of the power of a lens. The unit universally adopted
is that of a lens of one metre focal length, which is called
a dioptre, and is denoted by the symbol D. We know that a
convex lens of 25 cm. focal length is four times stronger
than one of 100 cm. focal length. This is very simply
expressed by calling the former lens +4D and the latter
LENSES 81
+ 1D. Note that it is the first focal length, /', that is
considered and that determines the sign of the lens. We
give an example or two, so as to make this nomenclature
quite clear.
What is the power of the following lenses in dioptres ?
two concave lenses of 10 cm. and 20 cm. focal length
respectively, and one convex lens of 80 cm. and another
convex lens of 22*5 inches focal length.
The first focal distance of a concave lens is negative,
so in the first case /' = 10 cm., or ^ metre; y, is
then 10D ; in the second case/' = 20 cm. =  5 metre,
so , is 5D.
The first focal distance of a convex lens is positive, so
if /' = 80 cm., the power of the lens in dioptres is J^f, or
+ 125D.
In every case the dioptric power is given by j, in metres,
100
or jr in centimetres. When we have /' given in inches,
we must convert it into metres. As 39*37... inches are
22 '5
equivalent to one metre, 22 '5 inches are equivalent to QQ . Qf 
t>y 'of
1 39*37
metres, so^,in metres is = 1*75. The dioptric power
J Z&'d
of this last glass is then +175D.
Thin Lenses in Juxtaposition. A succession of thin lenses,
such as that shown in Fig. 36, has practically the same effect
as that of one lens whose power is represented by the sum of
the dioptric strengths of the components of the system. Let
us suppose that the first lens (a meniscus) is I ID, the second
lens H10D, and the third lens 9*5D; the sum of the
dioptric strengths is +1 f 10 9*5, or +1'5D. Conse
quently, the system will be practically equivalent to one
lens of 4l'5D. This illustrates the extreme simplicity
that results from denoting lenses by their power instead of
by their focal length. If the thickness of the lenses, or of
O
82
GEOMETKICAL OPTICS
their combination, had to be taken into account, a correction
would have to be added, which will be explained when
dealing with cardinal points.
FIG. 36.
When a space separates the several lenses of the system,
the rule just stated does not apply ; for such a case the.
reader is referred to p. 91.
Optical Centre. We have frequently made use of the
term " the centre of a lens," and the reader most probably
,o thinks that it is equivalent to the mid
\ point of the thickness of the lens. It is
so in symmetrical biconvex or biconcave
lenses, but it may be outside the lens
altogether, as in Fig. 37. The optical
centre of a lens may be defined as that
point in which the line joining the ex
tremities of any parallel radii of the two
bounding surfaces cuts the axis.
In Fig. 37, BACiC 2 is the axis, and
and C 2 J 2 are two parallel radii.
FIG. 37.
The line joining J 2 and J b if produced, cuts the axis in 0,
which is then by the definition the optical centre.
The point is a fixed point, the position of which
depends only on the lengths of the radii and the thickness
t of the lens ; for by similar triangles
Also
C 2 O
ri  Cl Tl ~ A  A
^  0,0 = r a OB ~ OB
OB  OA AB = r 2 
"OB OB r a
OPTICAL CENTEE 83
So OB= ^ and OA = ^~ (a)
r 2  n r 2  T*!
Consequently, in biconvex and biconcave lenses, when either
TI or r 2 is negative, r 2  n > r^ so AB > OB ; in other
words, is within the lens. The optical centre has the
following important property : any incident ray, such as PJ 1}
passing through the lens so that its direction while within
the lens passes (either actually or virtually) through the
centre will on emerging from the lens, have a direction
J 2 Q parallel to its direction PJ X when incident ; and, con
versely, if any emergent ray be parallel to its corresponding
ray, it must, while within the lens, have assumed a direction
that would pass through the optical centre. This property
of the optical centre follows at once from the fact that the
tangents at the two points where refraction takes place are
parallel, and therefore the effect on this ray is the same as that
due to refraction through a plate (p. 40), i.e. that the angle of
emergence is always equal to the angle of incidence 0. Pencils
that pass either actually or virtually through the optical
centre are called centric pencils.
It is an easy matter to find the optical centre of any
lens from the expression (a), if we are given the thickness
t, (AB), and the radii of the two surfaces. It is quite
immaterial which way the light is supposed to be travelling,
whether from P to Q or from Q to P (Fig. 38). In the case
of a biconvex or a biconcave lens, the ray while within the
lens actually (not virtually) does pass through 0.
Let us consider the left surface as A, the first surface,
then if T! = 3, r 2 = 4, and t = 2, by (a) we know that
and 
r a  7*1  4  3 7
and we notice that is situated within the lens and nearer
the most strongly curved surface. Now, an incident ray PK'
that emerges in a parallel direction, as K"Q, will not follow
the course indicated in Fig. 38 within the lens, for it will be
84 GEOMETRICAL OPTICS
bent at the first surface, then pursue a straight course
through 0, and then be again bent on emergence in the
direction K"Q. It is, however, very convenient for purposes
of calculation to find two points, K' and K", on the axis
which will save us the trouble of calculating the angle of
obliquity of the path within the lens. This can clearly be
done by continuing the incident and emergent rays until
they intersect the axis in K' and K", or the nodal points.
This is a simple geometrical method, but it does not readily
lead to an analytical expression, and the accuracy of the
result obtained by a geometrical method depends, of course,
upon the accuracy and size of the drawing. There are, how
ever, two analytical methods for locating the nodal points
FIG. 38.
that we shall presently give, but we shall first deal with a
case in which the nodal points coincide at the optical centre.
* Refraction of a Sphere. Coddington Lens. The sphere
may be considered as a kind of double convex lens, and
there are certain advantages attending its use, which we
shall investigate.
Note, first, that all pencils (even those that are oblique)
which traverse the centre of the sphere pass normally into
it, so that in this case all centric pencils are also axial.
It may be readily seen, from (a), p. 83, that the geo
metrical centre of the sphere is also the optical centre of
the lens ; so it will be convenient to find an expression for
the focal distances when considered as measured towards the
centre.
SPHERE
85
Let (Fig. 39) denote the centre of the sphere, and let
P denote an object in front of it, Qi its image due to
refraction at the surface A, and Q the final image due to the
refraction of Qi at the second surface, B.
Let P denote PO = PA  OA = p  r lt
and q r QiO = Q X A  OA = ^  r b
and let Q QO = QB  OB = q  r 2 .
By the formula of p. 53,
or
Dividing by p'
Now regarding Q 1 as the object for refraction at the
second surface B, we note that p is to be replaced by
QiB = QiO + OB = q' + ra, and  by ^? in the formula
to M
Pr __!* P
So
or
or dividing by fiQ^'r^
86 GEOMETEICAL OPTICS
On subtracting (a)
J^L + J^I+l
u.g TI r
 <
112
Noting that r a = TV we see that  = , and by
r 2 n r 2
making P infinite we can find /"; similarly by making Q
infinite we can find/'. "We have then
i = ^v.2 and i.y^.l
/ A* r a / /*' r 2
and so we may write (5) as
111 /' /"
 or = 1
This tedious work might have been avoided had we been
able to use the method of " Cardinal Points," but it is well
to see the labour that is involved even in the simple case of
a sphere, if we are ignorant of the better method that we
shall shortly describe. It has been taken as an instance in
f f"
which our old formula 4. = 1 can be used if all the
P <1
distances involved are measured towards a certain point, in
this case the centre of the sphere. The section on Cardinal
Points will show that in any system, however complex, two
points can be found such that, if the appropriate distances
/' /"
are measured towards them, the formula + = 1 will
P 3.
hold good. In the case of a sphere these two points are
coincident.
The Coddington lens, represented in Fig. 40, is a very
convenient form of pocket magnifying glass, and has this
important advantage over an ordinary convex lens the
peripheral parts of the virtual image are as distinct as the
NODAL POINTS 87
central parts, provided that there is a central stop, so that
none but centric rays traverse it. Usually a deep equatorial
groove is ground in the lens, as indicated by the shaded part
in the figure. In practice it is found that the central aperture
must not be greater than a fifth of the focal length of the
lens. The defects of the lens are : (1) the image is curved
as the peripheral parts of the object are further away from
than the central parts ; (2) the working distance is very short ;
and (3) the field of view is limited ; as only those emergent
pencils which can enter the pupil of the eye are effective ; it
is advisable, therefore, to bring the lens as close to the eye as
possible.
The Stanhope lens is somewhat similar to the Coddington ;
it is a short glass cylinder with its ends ground convex to an
FIG. 40.
unequal degree of curvature. The object is placed on the
surface of lesser curvature, and the length of the cylinder is
such that when the more convex surface is turned towards
the eye, a distinct magnified image of the object is seen.
Nodal Points of a Thick Lens. The following method is
applicable only to lenses, and can, further, only be used
when the initial and final media are the same. Should they
be different, as in the case of the eye, it will be necessary to
use the method described later (p. 91), which is perfectly
general The only advantage of this method is its suitability
in the case of thick lenses, if we forget the formulae for
Cardinal Points.
Given the thick lens, we first find the position of by
(a), p. 83. We then find the position of the image of as
viewed from the incident surface (A), and call it K', the
first nodal point. We then find the position of the image of
when viewed from the other surface (B), and call it K",
88 GEOMETKICAL OPTICS
the second nodal point. A warning must be given when
dealing with menisci, as the point must always be con
sidered as situated in the glass, even when it is not, as in
Fig. 37, or totally erroneous results will be obtained. Con
sequently the method must be regarded simply as a trick or
device by which it can be mathematically proved that accu
rate results will be obtained.
We will take the case of Fig. 38 as an example, where
n = 3, r<i = 4, t = 2, and OA and OB we have found to be
f and f respectively. We are now considering the left
hand surface to be the incident surface A, so that light is
supposed to be travelling from Q to P. The problem is the
same as that on p. 57 ; we first find the image of formed
by the surface A. As is in the glass, ju = 1*5 and // = 1,
so
Pfi" f(6) 36
\T' A /VM ft _ tV * __ i ^ _ {_ __ _ _ _
r 2  p ./"'" _  9 ~ 69
Similarly for the surface B*
,,
/ 
and /." .J;
And as OB or p'
12
23
2" 92 23
The position of the nodal points is quite independent of
the direction in which the light is travelling; it is only
their names (first or second) that are changed. We have
been considering light passing from Q to P, so K' refers to
the lefthand surface. In the diagram the reverse condition
is indicated ; the light is passing from P to Q, so that the
righthand surface is the incident surface, and the nodal
SPHEKICAL ABEEEATION
89
point corresponding to it is called K', while K" refers to the
emergent (left) surface.
The mode of using these nodal points will be illustrated
later, when we deal with the geometrical construction of the
image of a complex system.
Spherical Aberration. In Fig. 41 a beam of parallel rays
is shown that encounters a double prism ; the more central
rays SI and ST intersect at E after traversing the prism,
FIG. 41.
FIG. 42.
while the more peripheral rays PJ and P'J' intersect at a
more distant point, L. Clearly the aberration EL might be,
obviated by bevelling the peripheral parts of the prism, so
that the incident rays at J and J' would undergo a greater
deviation. In a spherical lens (Fig. 42) this bevelling is
90 GEOMETEICAL OPTICS
carried out so that it acts like a prism, the strength of which
is continually increasing from the axis to the periphery.
Unfortunately, a spherical surface is not quite the right
shape indeed, the bevelling has been carried too far, for the
peripheral parts of an incident parallel beam intersect at a
closer point than the more central parts. The diagram
illustrates what is called spherical aberration (undercor
rected), while Fig. 41 will indicate what is called overcor
rection of spherical aberration.
There is one obvious way of making the focus of such a
lens more definite, i.e. by cutting off the aberrant peripheral
rays with a stop, so that the focus is only formed by the
intersection of the thin axial pencil. Another method which
has certain advantages, as has been pointed out by Lord
Eayleigh, is to block out the central rays and use only the
peripheral ones. This, however, has been rarely used in
practice.
Just as with reflection at a mirror, when an oblique or
eccentric incident pencil is considered, the refracted pencil is
astigmatic, and presents the same focal lines with the same
sphenoid shape between them. All this has been omitted in
the diagram for simplicity. The short line on the axis
between the focus of the peripheral and that of the more
central rays may be regarded as the secondary focal line,
while the intersection of the peripheral with the central rays
indicates the position of the primary focal line sagittal to
the plane of the paper.
It may reasonably be asked, Why are lenses made of this
erroneous shape? The answer is that it is impossible to
mould glass of the right shape with any approach to accu
racy, and grinding by hand to any shape but spherical is a
most laborious and difficult undertaking.
If we confine our attention to thin oblique pencils, we
see that they may be of two kinds : (1) an oblique eccentric
pencil that is incident upon a peripheral or eccentric part of
the lens ; and (2) an oblique centric pencil that traverses
(either actually or virtually) the optical centre of the lens,
as PQ in Fig. 38 or PQ in Fig. 37. The exact mathe
CAEDINAL POINTS 91
inatical investigation of the form of pencil after (1) oblique
eccentric refraction through a lens is most tedious and diffi
cult, and does not lead to any simple approximate expres
sion ; and, further, it is of little practical importance, except
to certain instrument makers, so we must refer the scientific
mathematician to more advanced treatises on this point.
(2) Oblique centric refraction will be briefly treated in the
Appendix (p. 121), as it is of considerable practical importance.
"Cardinal Points. Gauss has shown us how to extend the
f f"
use of the formulae 4 = 1, etc., not only to thick
lenses, but also to any refracting system however complex,
formed of any number of media bounded by centred spherical
surfaces. The only requisite is to find the position of two
points called the Principal Points of the system under
consideration. When the distance of the First Principal
Point H' from the object P is denoted by PH', and the
distance of the Second Principal Point H" from the object Q
TjVTTf ~F"TT"
is denoted by QH", we have ^w f nTT ,, = 1 universally
true, where F'H' is the distance of H' from the First Principal
Focus F', and where F"H" is the distance of H" from the
Second Principal Focus F" (Fig. 44).
Before actually dealing with the problem, we will show
exactly what it is that we want to find. Let HiA and H 2 B
represent the principal planes of two thin concave lenses
(Fig. 43). An incident ray parallel to the axis will, on
traversing the concave lens at HI, be refracted in the
direction F!^ as if it were proceeding from the second
focus of the first lens FI ; on now meeting the second lens
at H 2 , it will again be deviated in the direction H 2 R as if it
were proceeding from K We wish to find a lens which shall
have an equivalent action to these two lenses. It is clear
from the diagram that a concave lens placed in the position
KX, if its second focus be at NI, will have precisely the same
effect on the incident ray SHiK as the combination of lenses
had on SE^. The whole object of Gauss's method is to find
the position of X, i.e. the situation of the equivalent lens,
92
GEOMETRICAL OPTICS
and also its focal distance NX. In such a simple case as
this, it is an easy problem to solve by purely geometrical
means, but Gauss has shown us how to deal with any system,
however complex.
N A
FIG. 43.
The method used by Gauss is intricate and involves a
considerable knowledge of mathematics, but as we know the
result of his calculations, we shall be able to find the positions
of H' and H" by easy algebra if we treat the question in the
right way. In any algebraic problem it is necessary to pay
the utmost attention to the algebraic statement of the problem,
but after it is once correctly stated, think no more about
the meaning of the future operations until you get your result.
" Put it into the algebraic mill and turn the handle." The
chief difficulty of all beginners in mathematics is that they
try and think what each algebraic step means. This is quite
unnecessary; thought is only required when stating the
problem.
We will take as an example the thick lens in Fig. 44,
and use the following symbols to denote the various distances.
The thickness of the lens AB = t, the distances of A
from the first (fi') and second (fi") foci, due to the first
refraction at A are f x 'A, =/i' and f/'A =/i", and similarly
when f 2 ' and f a " denote the first and the second foci due to
the second refraction at B, f a 'B =/ 2 ', and f 2 "B =/ 2 ". The
values of these symbols can be determined in any given case
CARDINAL POINTS 93
by the wellknown formulae of p. 55 ; they are indicated in
the special case illustrated in the diagram by the letters
below the line, whereas H' and H" that we have to find are
shown above the line. We shall also use the symbols ti and
h" to denote the distances H'A and H"B respectively.
Now, if P denote any object (not illustrated in the
diagram), PA or p will denote the distance of the surface A
Stff
from it, and we can easily find the position of the image Qi,
due to the first refraction, for if q\ denote the distance QiA,
g/i '
P  /i
Kegarding now Q! as the object that forms the final image
Q by refraction at the second surface B, we can as easily
express QiB in terms of QB or q. For
QlB = ^7?
But QiB = QiA f AB = qi 4 t, so we can eliminate qi by
substituting for it our previous expression. We have then
We have to find the positions of H' and H" so that the
formula
FH' F'H"
shall give~a result that is identical with that given by (a).
Now, ' PH' = PA f AH', or PA  H'A, i.e. p  ti ;
similarly, QH" = QB  H"B = q  &", and if we denote
94 GEOMETEICAL OPTICS
FH' by F' and F"H" by F", this last formula can be
written
All, then, that we have to do is to make these two formulae
(a) and (b) identical with each other. For (a) we shall
write
or
pq(t +/r / a ') yf/i'/t" + ft") + gffi'ft  ft 1 ) + *fW = o (O
And for (b) we shall write
F'q  F'b" f F"p  F"h' = pq ph"  qh' + k'h"
or
^ + ^') + ^'^" 4. w 4. jr%' = . (V)
On comparing the coefficients of (a 1 ) and (5') we see that
the two expressions will be identical if
/ 4_ / " _ / ' fi'f*" + (/a" __ ^/i' fif* __ _ (/i/a" _
72 A" + ^"' 1^4^' ~h'h" + F'h" + F"ti
or calling this expression K,
,
A
and if ^'A" + F'h" + F"h' =
// /* / f ' ' f "f '
.... AA ,.  , if V $, if ^ = , and if F" = fl 
these values are clearly consistent, and they are therefore
the solutions required.
In the case illustrated in the diagram (Fig. 44), the
radius of the first surface, TI = 4, that of the second sur
face r 2 = 2, fjiQ = 1, u = 1/5, and t = 3.
CARDINAL POINTS 95
/'or ~
./i or ,
JJL fiQ
5
and K = t +ff ft = 3 + 12  6 = 9
fit 8x3
^ = K= 9~ :
j\, 7
ft 4x3
, fi't 8x3 M .
.A=y 9 = 2,
If in this case we were to use the formula on p. 70,
which neglects the thickness of the lens, we should have
8
or /" =  8, and /' or /" = 8, a result that is quite
erroneous. However, on adding the correcting term
  , or on using the form


r = r2 " ri " p) we get r = i 3
.'./" or /' = 5j, which is correct.
When two compound systems, A and B, are combined,
A being the first system traversed by the incident
light, t = Ha'H/, i.e. the distance of the first principal
point of the second system (B) from the second principal
point of the first system (A) ; whereas ti = H'H a ', or the
96 GEOMETRICAL OPTICS
distance of the first principal point of the complete system
measured towards the first principal point of A. Simi
larly A" = H a "H 6 " and K= H a "H 6 ' +/"/*', the sub
scripts a and b denoting the systems to which the letters
refer.
These formulae are a little difficult to remember, but
when reference can be made to them they are easy to
employ, and they are of universal application. We shall
give three examples of their use, to illustrate their simplicity
and value.
Ex. (1). The ordinary form of eyepiece for a microscope
(Huygenian) consists of two convex lenses, the distal one
being called the fieldglass, and the proximal one, to which
the eye is applied, the eyeglass. It is found that a No. III.
eyepiece, with an alleged magnification of eight diameters,
cannot be used as a magnifying glass when held in the
normal position before a microscope slide. Explain this, and
show how it can be used as a magnifying glass.
In all Huygenian eyepieces the fieldglass (A) has a focal
length three times that of the eyeglass (B), and the distance
between them is onehalf of the sum of the focal distances.
The advantages obtained by this construction cannot be
explained in this place, but it may be stated that the
spherical aberration of the system is less when each lens has
its second focus at the same point, and that the size (though
not the position) of the image is achromatized when
t = K/.' +/,')
With a No. III. eyepiece f a ' = f ins., and // = f in.
As we may regard the lenses of negligible thickness, we
have t =  ins.,
and *<**+/."/' = 444 = f
ET I Jajb
= i^r == ~
xs
^=*fj "* and ^"= 5
HUYGENIAN EYEPIECE 97
The position of the cardinal points of this Huygenian
eyepiece is illustrated in Fig. 45, and it will be seen that the
first principal focus (F') is
within the eyepiece; and
we know from p. 78 that
the object must be placed
not further off than F, so
clearly the eyepiece will
not act as a magnifying
glass when used in this FIQ
way. However, F" is situ
ated outside the lens, beyond B, so that if we reverse the
eyepiece, using A as the eyeglass and putting the microscope
slide in the position of F", it will form a very efficient mag
nifying glass. In such a position, when the eyepiece is
reversed, F" and F' are simply interchanged. We have
shown that the magnifying power of a lens is measured by
^. In this case 4=10 ^^ = 8.
When a Huygenian eyepiece is used in a microscope, the
image formed by the objective of the instrument is formed
within the eyepiece at F', so that magnification results
normally.
If the two lenses of the eyepiece were placed in contact,
the power of the combination would be
r+pf+fl
Ja Jb
or the combination would then be equivalent to a lens of  in.
focal length, while its magnifying power would be doubled,
being 16 ; it would, however, manifest all the chromatic and
aberrational errors that the Huygenian eyepiece in part corrects.
Ex. (2). Suppose that the eyeglass in a No. III. Huy
genian eyepiece were replaced by a concave lens of equal
strength : what would be the power of the combination, and
what use could be made of the instrument ?
Here/,' =  and K 01 t +/."/.' J_f +  =

00
and as F' = *& = m , the power = 
A Jf
98
GEOMETKICAL OPTICS
The power of the combination is therefore 0, i.e. all inci
dent parallel light emerges parallel, and it might be thought
to be useless as an instrument; but a little consideration
will show that it will act as a very efficient telescope
indeed, this is the form of the Galilean telescope or opera
glass.
A very distant object, subtending a visual angle 9, will
also make the angle f a "AD at the fieldglass (Fig. 46) equal
FIG. 46.
to 8, and an image of the object would be formed in the
second focal plane at f "D. But f a "D is also the situation of
the first focal plane of B, consequently the incident parallel
light which is converging towards D will, owing to the inter
position of the concave lens at B, proceed as a beam of
parallel rays in the direction BD (the spaced and dotted
lines). Similarly, the axial incident rays which, after
traversing A, tend to converge towards f a " will, owing to the
action of B, proceed as a parallel beam in the direction of the
axis. Note, however, that the direction of the oblique pencil
has been changed, as it now makes an angle 0' with the axis
instead of 0. An eye, therefore, if adjusted for parallel rays,
placed close to B will see an erect virtual image subtending
the angle 0', and the magnification of the instrument will
be
tanj?' _f a "A_ 5 . 5
tan 6 ~ 'B = "2 * 6
CARDINAL POINTS OF THE EYE 99
Ex. (3). We will now find the cardinal points of the
human eye. According to Tscherning's most recent investi
gations, the lens of the eye has a focal length (0) of 51/34
mm., and its principal points, indicated as H! and H a in
Fig. 47, are so situated that HiAi = 2308 mm., and H a A 2
= 1*385 mm. The thickness of the lens is 3*9 mm., and it
is placed 3*6 mm. behind the cornea (Ao). We shall con
sider the media to be of the same refractive index 1*3375,
bounded by the cornea of radius  7'8 mm. In the diagram
FIG. 47.
the incident light is presumed to be travelling from right to
left, so we will regard this as the positive direction.
Taking first the corneal refraction, we find
and /"= fif'= 3091 mm.
The distance t or
AoHi = AoAi + A^ = 36 + 2308 = 5*908 mm.
and j^or t + /"  f = 5908  3091  51*34 = 76'342
IT' A  T.' /'< 23>11 * 5908 1 7QQ
.. H Ao, ^.e. h or  = . =  1'789 mm.
100 GEOMETRICAL OPTICS
 2311x5134
1 OT F = 4 =  _  =1564 mm.
F'H" or J"  =  =  2079 nan.
The distance of the second principal point from the cornea
can be easily obtained, for
H"A = H"H 2 + H 2 A 2 + A^ + A^o,
or H"Ao = 3973 + 1'385  39  3'6 =  2142 mm.
The point H" is therefore only 0'353 mm. from H'. We
shall not, then, introduce any appreciable error in consider
ing that they coincide in one point H, towards which both
the focal distances are measured.
Nodal Points of a Complex System. When we have found
the cardinal points by the above method, it is a simple
matter to find the nodal points. Take any point S in the
first focal plane (Fig. 48), and through it draw S JiJ 2 parallel
to the axis; join J 2 F". All light proceeding from S will
emerge from the system in a direction parallel to J 2 F", e.g.
SIiI 2 K. From S draw the ray SDiK' parallel to J 2 F", repre
senting such a ray, and from D 2 the point on the second
principal plane corresponding to D b draw D 2 K"E parallel
to J 2 F", cutting the axis in K". Then K' and K" are the
first and second nodal points of the system. We shall see
the use of them in the next section. Meanwhile we will
devote a little attention to the diagram. We notice first that
every ray incident on the first principal plane travels parallel
to the axis until it meets the second principal plane ; this is
a characteristic property of these planes, and as any object
in one plane, when viewed from the other side, will be seen
without any alteration in size, they are often called planes of
unit magnification, or the Unit Planes.
As the sides of the A F'SK' are parallel to the sides of
the AH"J 2 F", and as F'S = H"J 2 ,
K'F = F"H"or I?".
Also the A EF"K" = ADiJiS, as the sides are parallel, and
as EF" = D 2 J 2 = D^!,
/. K"F" = SJ X = F'H' or F'
NODAL POINTS
101
It is also clear that K'K" = DiD 2 = H'H",
and that K"H" = K'H' = F'H" + F'H' = F" + F.
FIG. 48.
We have found then all the cardinal points of the standard
emmetropic eye which are given in the subjoined table.
CARDINAL POINTS OP THE EMMETROPIC EYE.
H'A . .
H"A . .
. .  1789 mm
. . 2142 mm.
H'H" \
K'K" /
. . . . 0353 mm.
K'A
K"A . .
. . 704 mm.
. .  739 mm.
H'K' \
H"K"/
. . . . 525 mm.
F'A . .
F"A . .
. . 1375 mm.
. .  2293 mm.
F'H" \
K"F" /
. . . . 1554 mm.
F"H" \
K'F' /
. . . .2079 mm.
In every system in which the initial and the final media
have the same refractive index (i.e. when F"H" = H'F) the
nodal points K',K" coincide with the principal points H',H".
We see also that if we regard the principal points as coin
cident in the human eye in H, the nodal points coincide too
in one point K, and the distance of this nodal point to the
102 GEOMETKICAL OPTICS
principal point, or KH, = F"H" + F'H' =  2079 + 15'54
= 5 '25 mm.
Geometrical Construction of the Image. To find the image
of AB (Fig. 49), we first mark the position of the cardinal
points on the axis. We then join AK', and draw K"# parallel
to it. Then we can either draw AJ" parallel to the axis and
J"F"a through F", or we can draw AFT through F', and IT'a
parallel to the axis. In either case, a is the point of inter
section of the line with that drawn through the nodal point.
. i . HT _ F^H' _ F'H' F'
~ BA ~ F'B ~ F'H'BH' " T r ^
1 la F"6 _ F'H"  6H" _ F"  g
o m H"J" ~ F"H" ~ F"H" F "
On p. 75 it was shown that when thin lenses are con
sidered  = , and we see from Fig. 49 that  = ^^, so
op o .bJi.
that the nodal points of a complex system play the part of
the optical centre (0) in a thin lens. The distances &K" and
BK' are usually denoted by g" and g' in the books.
In the simplified schematic eye described above, the nodal
point K acts as if it were the centre of a convex spherical
surface of radius 5*25 mm. that forms the boundary of a
medium of refractive index
F" 2079
~ F' r iF54~ 1338
The tangent of the visual angle subtended at the nodal
point of the eye (p. 77) by an object is the same as the
tangent of the angle subtended by the retinal image at K.
If this retinal image be i mm. in height,
tan =
It is often simpler, when the object is very distant, to con
sider an optical instrument and the eye as forming one
complex system, as in that case tan is practically constant,
and if K" denote the second Nodal Point of the system and
K that of the unaided eye, it is clear that the magnification
of the system is as 6K" : &K
CONSTKUCTION OF THE IMAGE
103
104 GEOMETKICAL OPTICS
As an example, we may explain the action of a Baden
Powell lens, which is used as a handy pocket opera glass.
It is simply a convex lens of weak power which is held at
arm's length, and distant objects are viewed through it;
these then appear larger to the eye, or, when less correctly
expressed, they are said to seem nearer.
We will suppose that the glass is + 0*5D where /'
= 2000 mm., and that it is held at a distance of 500 mm.,
or about 20 inches, from the eye.
On considering the lens and the eye as one complex
system, and finding the cardinal points, we get the following
values :
K= t +f _ F' = 500  2000  1554 = 151554
,, ft 2000 x 500
" * r K = 151554 = ~
, F"t 2079 x 500 = 6859 mm.
h Or ^r = 151554
, _ /'J __ 2000x1554
K 151554 ~ *
p ,,^f"F" 2000(2079)^
~K~ 151554
It is clear, then, that the first principal point and the
first focus are both more than 100 mm. behind the eye.
This is of no importance in the problem now before us ; but
the second principal focus must be situated on the retina if
the object is to be seen distinctly, so we must find whether
this is so.
F" is situated 27'436 mm. behind H", which is 6*859 mm.
in front of the eye, consequently F" is 20*577 mm. behind
the principal plane of the eye. As in the standard eye F" is
20*79 mm. behind its principal plane, we see that, unless the
eye be 0*213 mm. too short, a sharp image will not be formed
on the retina.
Now, few eyes are of exactly the standard length, and
we will suppose that the eye considered is too short by
at least this small amount (less than 075D of hyperme
tropia), so that 6K" = F"K" = F'.
GRAPHIC METHOD 105
The magnification (M) will be as F"K" : F"K
F"K" F' 2051
= F'K = ^F' = T5^54~ 13
Consequently, the distant object will appear larger by
nearly a third. If the eye be myopic or too long, the Baden
Powell lens is useless without some contrivance to make the
image definite. The simplest effective contrivance is a card
with a pinhole in it held close to the eye. In this way the
circles of confusion on the retina are made much smaller, so
that the image .may be regarded as sharply defined.
Graphic Method for Cardinal Points. Professor Sampson
has devised a most ingenious method of finding the cardinal
points of a thick lens by a graphic method. It is an exten
sion of the graphic methods which we have frequently illus
trated in the preceding sections.
The diagram (Fig. 50) shows this method applied to the
case of the meniscus we discussed on p. 95. The thickness
of the lens AB is 3, so we measure AB in the positive direc
tion (upwards) 3 units. As A indicates the first surface, we
mark off fi'A in the negative direction to represent  8 units,
and fi"A in the positive direction to represent 12 units.
Similarly, dealing with the second surface B, we make f 2 'B
equal to +6 units, and f 2 "B equal to 4 units. The paral
lelogram referring to the first surface (A) is completed at KI,
and that referring to B is completed at K 2 .
Join KxK 2 , and produce to meet the horizontal line
through B at H", cutting the horizontal line through A at H'.
Then H'A is ti, the distance of the first surface from H',
and H"B is /*/', the distance of the second surface from H". In
this case they are both measured from right to left, so they
are both negative : H'A =  2f , and H"B =  1.
Join KI and f 2 ', and produce to meet the horizontal line
through A in F'; and join K 2 fi", and produce to meet the
horizontal line through B in F".
Then F'H' is positive and is found to measure 5 J units,
while F"H" is found to measure  5 J units. It will be found
that F'H' and F"H" correctly indicate the two principal
focal distances of the thick lens illustrated in Fig. 44.
106
GEOMETRICAL OPTICS
The method is of delightful simplicity and presents no
difficulty whatever, if due attention be paid to the directions
of the positive and negative measurements, and if it be noted
that the points fi" and f 2 ' must be in the same straight line
f A
\
1
\
7
\
\
'
\
/
\
\
I
\
\
1
Si
\ A 2
\ ' x
\/ \
/\
/ X  X
\
/ > \
/ 'S;\
s#
/ ^^
IS
FIG. 50.
as AB, and that K x is joined to f 2 ' (subscript 2), while K 2 is
joined to fi" (subscript 1).
To show the generality of the construction, it will be
sufficient here to notice that
f^'fa' = fi"A + AB  f 2 'B = t +/i" &+*,
and that from the similar triangles in the figure
FA f 2 f A or FH^+ H[A = fa f B__AB
or
, .. . F'B fj"B
Similarly, ~ fl = T^nr,
K 2 i 2 ii i2
LENSES 107
F'H" + H"B ri "A + AB
or
or  (Cf.p.94)
/2
(The above is not given as a proof of the construction, for
clearly FA = FTK' + K'A, and F'B = F'K" + K"B, or indeed
the sum of any other lines with the same end points. It
will be found that if the final medium has a different re
fractive index from that of the initial medium, the points
marked H' and H" really denote the two nodal points K'
and K" of the system. As from p. 101 we know that when
F" = F', the nodal points coincide with the principal
points, we may regard the points H' and H" determined by
this construction as always denoting the nodal points.)
QUESTIONS.
(1) The focal length of a convex lens is 6 ins. ; an object is placed
36 ins. from it. What is the relative size of the image, and where is
it formed ?
(2) Show that/'/" =(/'  p)(f" q) in all cases.
(3) The radius of curvature of the first surface of a thin converging
lens is 6 ins. If its focal length be 10 ins., and if the index of refrac
tion be 152, what is the radius of curvature of the other surface ?
What would its focal length be when placed in a tank of water 0* = ) ?
(4) A convex lens of focal length i in. is used as a magnifying
glass. The nearest point of distinct vision is 10 ins. from the nodal
point of the eye. Find the magnifying power (i.) when the lens is in.
from the nodal point, (ii.) when it is 1^ in. from the nodal point ; and
(iii.) when the object is i in. from the lens, and the eye is adapted for
distance.
(5) The cardinal points of the following thick biconvex lens are
required where r r = 4, r 2 = 2, t 3, and /x = T5. Use the graphic
method, and test your results by the algebraic formulae for the position
of the cardinal points.
APPENDIX
THE following notes are intended for those who require some
further knowledge of the subject which the previous elementary
treatment rendered impossible. At the same tune care has been
taken to include nothing which has no practical application ;
academic points of merely mathematical interest have been rigidly
excluded.
Oblique Reflection. Distances of the Focal Lines. We will
now give the proof of the formula quoted on p. 33, denoting
FIG. 51.
SP by u and F X P and F 2 P by v l and v 2 , the distances of the focal
lines due to the thin incident pencil PSQ (Fig. 51).
PCQ = OCQ  OOP = CSQ + SQC  (CSP + SPG)
= CSQ  CSP + JSQF,  JSPF,
2PCQ = 2PSQ + SQFi  SP^ / (a)
110 APPENDIX
But owing to the equality of the angles formed by the intersect
ing lines SP and F X Q, PSQ + SQF, = PFiQ + SPF W and on
substituting this expression in (a) we get
2PCQ = PSQ + PF,Q ..... (J)
Now, as the formula that we wish to find is only true for thin
pencils PSQ and PF X Q, we may substitute the chord PQ for the
arc PQ, and regard PSQ and PF X Q as triangles; moreover,
CQS or FjQC = <.
* r>Qn Sn sn
Then m A PSQ, gp  ETQS " E{90 3 +" CQS)
PQ = sin PSQ
'* SP ~ : cos <
andinAPFO PQ _ sin PF,Q _ sin PF,Q
m & W f* F X P ~ sin PQF, ~ sin (
PQ sin PF t Q
'''Ff~~ cos<j>
therefore in the limit when PSQ and PF X Q are very small
= . (c)
Note that as PQ is measured from below upwards, but SP
from right to left, the angles PSQ and PQS are measured in
reverse directions ; a similar precaution must be exercised in
dealing with the angles PFjQ and PQFj.
On substituting the expressions in (c) for those in (b), we can
write
PQ COS< , po  ^
.
u v r cos
Again, since A F 2 PC + A GPS =
sin <j> f ru sin <^> = V#JL sin
or, on dividing throughout by \v#u sin ^>,
1 1 __ sin 2< __ 2 sin < cos rf> __ 2 cos <
u v 2 ~~ r sin <^> "~ r sin < r
The reader may be inclined to think that the consideration
of such very thin pencils is of little practical use, but it must be
CAMERA LUCIDA
111
remembered that only very thin pencils can enter the pupil of
the eye, so that only thin pencils need be considered, if the
image formed by the mirror is viewed directly by the eye.
Camera Lucida. The Wollaston prism shown in Fig. 52 is
the usual form of camera lucida that is used in sketching objects
from nature. It is a glass
prism that presents four
angles, one of which is 90,
the opposite angle 135, and
the remaining two angles are
each 67J. Light from the
object enters one of the faces
normally, and is transmitted
without deviation to the
second face as SI; at I,
however, its angle of inci
dence is 67^, much greater
than the critical angle for
glass, so that it is totally
reflected at I as IR ; again
at R it undergoes total reflec
tion; and finally it emerges Q
normally to the upper sur FlQ 52>
face towards the eye at E.
The eye will therefore see a virtual image of the object pro
jected downwards in the direction EQ. In practice the eye
is placed over the edge near T, so that one half of the
pupil receives the light from the prism, while the other half is
viewing the sketchingblock and pencil placed below in the
neighbourhood of Q. It is most important that the image of
the object should be accurately projected on to the plane of the
sketchingblock, so the upper border at T is ground concave so
that it will have the effect of a 4D lens. In this way the
image is projected about 10 ins. from the prism, which will be
a convenient distance for the sketch to be made. The image
seen is erect, as there are two reflections ; had there been only
one reflection the image would have been upside down.
Oblique Refraction at Plane Surface. Focal Lines. We
will now consider the refraction of a thin oblique pencil at a
plane surface. If POQ (Fig. 53) be an incident oblique pencil
112
APPENDIX
originating from the source in a rare medium and refracted at
PQ in the direction RB', the primary and secondary (virtual)
focal lines will be formed at F x and F 2 . As the pencil considered
is very thin, we may substitute the angles POQ and PF X Q for
sin POQ and sin PI^Q, and we may disregard the difference
between the angles AOP and AOQ and consider that they are
FIG. 53.
each of them equal to <f>, and similarly we may consider F 2 as a
point, and consider that AF 2 P = AF 2 Q = <'.
If we denote OP by u, Ff by v lt and F 2 P by v& we have
in A POQ,
and in
PQ __ sin POQ __ sin POQ
OP "~ sin PQO "" coTAOQ
PQ
so
sn
POQ = ^ cos $ and PF X Q = ^ cos
Now,
sn
sn
cos
cos <f>
(a)
and we wish to find from Fig. 53 an expression for ,, or the
,. ... 
limiting value of
A
, i.e.
POQ
FOCAL LINES 113
When, therefore, POQ and PF t Q become infinitesimal,
PQ
d<f> u v l cos < p.' cos <

,AX
(A)
AP AP
And since sin < = and sm 9
/ sin <j> _ F 2 P
r sln~'  OP
(B)
The refracted pencil will be astigmatic, and a sphenoid will
be formed between the secondary and primary focal lines (F 2 and
Fj) exactly like that described on p. 65, when oblique refraction
at a concave surface was considered from a source of light whose
distance u was less than /u, (FjP). A blurred image of the
point will be formed at the position of the "circle of least
confusion " between F 2 and F l (as represented by D in Fig. 56,
where, however, F,>P is greater than FjP).
Oblique Refraction through a Plate. Let t denote the thick
ness of the plate, and let I denote the length of the path NM of
the thin pencil through the plate in Fig. 19. For the first
refraction we have v u . ^ and v 2 = ~u by (A) and (B).
^o cos j < [j,
For the second refraction at M the angle of incidence is <', and
that of emergence is <, while the relative index is , , so that on
replacing u by i;, + I or v z + I in the respective equations, and
making the other appropriate substitutions, we get
. (A')
..... (B')
The distance between the two foci, F 2  V, or^l 1 
fj. \ COS <p
114 APPENDIX
may be taken as a measure of the indistinctness of the image.
However oblique the pencil the distance separating the two foci
cannot exceed ~, I ; for direct pencils of course < = 0, and the
two foci are coincident at one and the same point distant u + tl
from the distal surface of the plate. If preferred we may sub
stitute t sec <' for I, but the form given above is easier to
remember.
When a small object S is seen obliquely through a glass plate
(Fig. 54) the following points may be observed :
(1) The image S' is
blurred, as it is represented
by the" circle of least con
fusion" between the two
focal lines.
(2) The upward dis
placement is greater than
when the object is viewed
normally, for in that case,
as we found on p. 40, the
upward displacement is
^ t, or onethird the thickness of the glass plate.
(3) There is in addition a lateral displacement.
Refraction at Spherical Surface. Focal Lines. The length
of Vi is determined in this way (Fig. 55). When the pencil is
thin, CPO or CQO may be represented by <, and CPF 2 or CQF
sn s sn
a A
_
sin (W^OQC) cos CQO
Similarly, in A F a PQ,
PQ _ sin PFjQ _ sin PF X Q _ sin
FjP ~~ sin PQF, sin (90  F a QC) cos CQF 2
Therefore in the limit when POQ and PFjQ are very small,
PQ C , PF lQ = PQ' and POQ = *
FOCAL LINES
Now, /AQ sin <f> = p.' sin <j>,
115
cos
cos
As we wish to find from the figure an expression for , , or
the limiting value of T~?>J we must no longer disregard the differ
ence between CQO and CPO and that between
Since the angles at the intersection L are equal,
i and CPF,.
or
or
FIG. 55.
POL + LPO = LCQ + CQL
POQ + <
And similarly, since the angles at K are equal,
PFjK + KPFj = KCQ + CQK
4
. A<A _
"
or in the limit
But
 PCQ
S <> 1
s $ _ 1\
v f)
116 APPENDIX
fl' COS 2 <' fl' COS <f> _ COS 2 <f> COS
r
\
 cos <f> )
V
sn < cos <' cos <j> sin <'
r sin <'
_ S inQ <#>')
OI  .  r t ~ > I .tV J
/*, w r sin <
Again, since
ACPO = ACPF 2 + AF 2 PO
Jrw sin < = Jr^ sin </>' + Jv 2 w sin (<#> <')
or on dividing throughout by fyruv z sin <^>'
sin < 1^1 sin > <'
' ~
sin <f> V 2 ~ u r sin
These formulae (A) and (B) are only true for extremely thin
pencils, but it is only such that can enter the pupil of the eye, so
that they are applicable in all the optical questions relating to
oblique vision in the human eye. They will also be used in the
next section when we deal with thin oblique centric pencils
traversing a lens.
It is evident from formula (A) and (B) that
 ( 1  sin A')  1  rin '* = JL _ 1
MtV
or !_
Ho V l U
fi' . V 2 sin 2
 (v.  Vl ) = ^
When yu, has the value 1, this expression takes the form
sn
and we see that when ^ 2 and ^ x carry the same sign, i.e. when
the two focal lines are on the same side of the refracting surface,
^! as w=
CIRCLE OF LEAST CONFUSION 117
On considering the caustic curve formed by a wide divergent
pencil issuing from a point source, it is clear that the cusp of the
caustic points away from the spherical surface when v z > v lt but
that it points towards the surface when v 2 < Vj. No caustic is
formed when v 2 = V 1 ; this can only occur when < = 0, or when
u = fj.v l ; the refracting surface must then be concave, for if
< = 0, u = r t and if u = pv^ as v must be positive, it must be
greater or less than u, according as /u, is less or greater than 1.
Circle of Least Confusion. It has been shown that if the
surface be not aplanatic for the source, whenever the incident
pencil is oblique the refracted or reflected pencil is astigmatic,
crossing itself in two focal lines and tracing out a sphenoid sur
face between them. It is required to find the position and size
of the smallest crosssection of this sphenoid surface represented
by DK in Fig. 56, which is practically a repetition of Fig. 15.
The surface (ab) at P is that part of the mirror or lens which
gives rise to the astigmatic pencil that has its focal lines at ^
and F 2 . For geometrical convenience squares have been described
on the radius (k) of the circle of least confusion, and on the
radius R of the effective aperture of the receiving instrument.
Now, DP (or x) and k, i.e. the distance and side of the small
square at D, can be easily expressed in terms of a and b. Note
that when comparing the vertical sides of the rectangles at P and
D, while a is measured upwards k is measured downwards, so
that they carry opposite signs.
a=~k or ~F '''^ = X ~1T ' ' (V)
F 2 P_F 2 D . v*_v z x
~b"~^k~ ' 0~ k (2)
On eliminating k from (!') and (2') we obtain
x v l .Vz x
a = 
which, on dividing by x, reduces to
118 APPENDIX
In practice, however, we have more often to consider the
cases in which a real image is formed at the back of an eye or at
the back of a photographic camera. It is, therefore, much more
convenient to express the distance (x) and the radius (k) of the
circle of least confusion in terms of the distance (s) of the
receiving instrument and the radius (R) of the stop used with it.
Let H' represent the first principal plane of the receiving
instrument (distant H'P or s), which only allows a pencil of
radius R to be transmitted through it ; if the receiving instru
ment be an eye, R represents the radius of an equivalent pupil
placed in the first principal plane of the eye.
By similar triangles we have
FT!! FT!) DF, . #1 s x v\ /, \
or
R k k R k
F 2 H F 2 D v 2  s # 2  x
and ~BT ~T ' ' ~IT T
i\ f #2 2s v 2
R(v } 2  v^
or k = V^  r . . (a)
 
Again, from (1) and (2) we also see that
k _ x v l __ v 2 x
R~~ Vi s ~" # 2 s
or x(v z  s)  vj)* 4 ^s = v v v z v<,s x(v v  s)
.. X(VL + v z 2s) = 2viV^ s(vi f v 2 )
2^sfa + t,)
Vi 4 ^2 2s
These two equations () and (b), determining the radius and
the distance of the circle of least confusion of an oblique pencil,
are always true whether reflection or refraction are considered.
If the receiving instrument be an eye, and the lens at P be a
spectacle glass worn in its appropriate place (the first focal plane
of the eye), a simple expression can be obtained for the radius r
of the retinal confusion circle formed by the retinal image of this
confusion circle at D.
For r F>
~
CIKCLE OF LEAST CONFUSION 119
120
APPENDIX
But as the lens is placed in the first focal plane, PH' = F',
so 8 = jT, and F' p = PH'  DH' = PD = x;
sR(v z  flQ
_ =
= ~~
 s(v! 4
Least Circle of Aberration. These formulas are of no use
when we are considering the spherical aberration of a lens
used with full aperture, as in Fig. 42. Let y (Fig. 57) denote
R
FIG. 57.
the semiaperture of a convex lens the focus of which is at F", and
let BTL denote an extreme ray ; it is clear that if a screen were
placed at F", there would be a bright point at F" which would be
surrounded by a halo of light extending to L and L'. The Jine
F"L is called the Lateral Aberration (I) of the extreme ray RTL,
while F"T or a is called the Longitudinal Aberration of RTL
which cuts the axis at an angle H"TR or \jr.
Clearly a = F'H"  TH" = f "  y cot ^,
and I = a tan \j/.
The point E represents the position of the first focal line (/j),
and the point T that of the second focal line (/ 2 ) of the extreme
ray RTL, and the curve EF"E' represents the Caustic Curve
formed by the lens when its effective aperture is R'R. It is
obvious that the light is concentrated over the smallest area at D,
which marks the site of the Least Circle of Aberration, its
radius DK or k is determined by the point where the extreme
ray RTL cuts the caustic E'F". It is usually stated that
F"D = a and that k = l
THIN OBLIQUE CENTRIC PENCILS 121
This is only a very rough approximation, as the position of D
can only be determined by tracing the caustic, 1 which may be of
almost any shape, some being long and narrow, while others are
short and stumpy, so that no general expression can be given for
the position of D. If in any case the position of D is given, of
course Tc = TD tan \j/.
Thin Lens. Thin Oblique Centric Pencils. It has been
already demonstrated that all oblique pencils which traverse the
optical centre of a lens emerge at an angle equal to that of
incidence (<j>) at the first surface, and that the angle of incidence
at the second surface is equal to the angle of refraction (<') at
the first surface. This fact is universally true, however different
the radii of curvatures of the respective surface may be.
Fig. 58 gives in a simple diagrammatic way the results of the
refraction of a thin oblique pencil PO that traverses the centre
O of a thin concave lens represented by the plane HO.
FIG. 58.
By the method of p. 116 we find the positions of p l and p. Z)
the primary and secondary focal lines due to refraction at the
first surface. Then, regarding p l as the virtual object for the
second surface, we find g 1 as the final primary focal line due to
this oblique centric refraction through the lens, and similarly
q represents the final secondary focal line.
1 A paper of mine in the Proceedings of the Optical Convention, 1912,
describes a method of tracing Caustic Curves.
I 2
122 APPENDIX
Hence, if r^ and r 2 represent the radii of the first and second
surfaces respectively, and if PO = U, pf> = v lt and p 2 O = # 2 , we
have for the first refraction
p! cos 2 $ _ cos 2 <ft __ // 1 __ sin (< <') . ^
U "~ A 2> 2 U " T sin <>'
For the second refraction <' is the angle of incidence, < is
that of refraction, and ? is the relative index, so if qf> = V^,
and q 2 O = Vjj, we have
x cos 2 < cos sin '  <>
Vi jw 2 v z r. 2 sn </>
cos 2 1  sin <> ~
sn
On adding (a) we obtain
1 1 sin 
o,
cos
This formula is, of course, only true for thin centric pencils ; it
is of no use when the whole aperture of a lens is employed to
form an image, but only when a diaphragm with a small central
perforation is used with the lens. It is, therefore, applicable to
the case of a camera lens when a small " stop " is used to make
the image more sharp, but perhaps the most important professional
use of it is in the case of spectacles. The pupil of the eye then
limits the width of the effective pencil, and when the visual lines
traverse the centre of the spectacles an application of the formula
gives rigidly accurate results. If the wearer gazes through
eccentric portions of his glasses, the results are only approximately
true, as then an investigation into the problem of oblique
eccentric refraction would become necessary. However, for all
practical purposes the simple formula we have obtained will
suffice.
Shortsighted persons are often seen wearing their pincenez
tilted on their nose, and they declare that they see better with
them in this position than when they are placed vertically. It
will always be found in such cases that the myope is astigmatic,
and that his pincenez do not correct his astigmatism. He is, in
TILTED LENSES 123
fact, making use of this property of tilted lenses to obtain an
astigmatic pencil which shall correct his error. I personally very
frequently order tilted spectacles to poor patients to whom the
expense of correct spherocylindrical lenses is prohibitive. Tilted
lenses are of use when the power of the spherical part of the
lens is high, and when only a weak cylindrical effect is required
in vertical meridian, i.e. when the plane axis of the cylinder is
horizontal ; they have this incidental advantage, that they are
lighter to wear.
We will take an example to illustrate this application of the
formula, as it will show the most convenient way in which such
questions may be treated.
Ex. A patient wears his distance pincenez ( 10D) tilted
30 from the vertical plane. Calculate the effect of this dis
placement.
As the object is presumed to be at a great distance, the
terms involving  vanish, and remembering that
when distances are measured in centimetres, we may write the
equation (A) as
D sin <<*>'
cos 2
 . 
I sin
where D stands for 10 dioptres, D l for the dioptric power of
the glass in the vertical meridian, and D 2 the dioptric power of
the glass in the horizontal meridian.
When < = 30 and /A = 1'5, D 2 is found to be very nearly
1095 dioptres. Consequently,
that is, the 10D glass with this tilt will act as if a cylindrical
glass of power 3*65D, with the plane axis of^the cylinder hori
zontal, had been added to a spherical lens of 1095D. If we
denote the cylindrical glass by D c we may find its power more
shortly thus
:.D, = K 1095) =5365D.
124 APPENDIX
Decentration of Lenses. It is sometimes said that the
refracting power of a lens is measured by the deviation that it
induces on a ray of incident light. On referring to Fig. 59 this
FIG. 59.
statement is seen to be untrue, or at any rate inadequate. The
First Focus of the convex lens is indicated by F', and the ray
F'H on emerging from the lens pursues the direction HS ; its
direction therefore undergoes a deviation F'HS' or HF'O, and it
is clear that the deviation HF'O depends upon the height of OH,
i.e. it depends upon which incident ray we choose to consider. It
will be noticed that the angle HF'O is negative, being measured
TTO /
in the clockwise direction, so tan HF'O = WQ or ji where HO is
denoted by /. The lens, considered only with reference to the
ray F'H has the same effect on it as a prism with its edge
upwards ; indeed, a lens may be regarded as a prism the strength
of which is continually increasing as one passes from its
centre O.
Now, if an eye were at S, it would see the image of an object
at F 7 displaced to a great distance in the direction of S'. The
lens would be said to be " decentred " HO millimetres downwards
(with respect to the eye). If the lower part of the lens were cut
away it would be called a Prismosphere, as it would act precisely
like a spherical lens which had been bisected equatorially with a
prism inserted between the two halves. Recently a great deal of
attention has been directed to the properties and applications of
decentred lenses or prismospheres, so that it may not be out
of place to give the very simple formula which connects the
decentration I of a lens with the deviation it induces.
Prism Dioptres, Centrads. The Prism Dioptre is a unit
proposed by Prentice, and is that angle whose tangent is 0*01 and
is symbolized by delta (A). Consequently, with a f 1 D lens of
PRISM DIOPTRES 125
which /' = 1 m. or 100 cm., when OH (Fig. 57) is 1 cm. a
deviation of lA is produced, for
In the prescription of spectacles the decentration, HO or I, is
always measured in millimetres, so that we have the simple
formula N = TQ A, where N denotes the number of prism
dioptres.
In order to obtain all the information possible from this
formula we must make some special convention about signs. A
prism, for instance, with its edge towards the right before the
right eye would cause divergence of the eyes, but if so set before
the left eye, it would cause convergence of the eyes. If, however,
we agree to call prisms with their edges upwards or outwards
(from the nose of the patient) negative, we shall get consistent
results. Further, if we regard decentration downwards and
inwards as also negative, the above simple formula will give us
full and complete information about the prismatic equivalent of
all decentred lenses. This convention about signs is a little
difficult to remember ; the memoria technica that I use is
Ztecentratiow, down and in, negative; Prisms opposite. For
instance, what effect will a 5D lens have if decentred 4 mm.
upwards ? This means what effect will a spectacle lens have if
its optical centre is displaced 4 mm. above the pupil of the eye ?
Here I = +4, it is positive because the decentration is
upwards, and
The decentration is therefore equivalent to a prism of 2 prism
dioptres with its edge set in the negative direction, and from the
memoria technica we know that must be either upwards or
outwards, and from the conditions of the question we know that
it must be upwards. The effect of this decentred lens will be the
same as that of a normally centred 5D combined with a prism
of 2 A set edge upwards.
There are, however, some objections to the prism dioptre as
a unit. It is not subject to the ordinary rules of arithmetic ;
126 APPENDIX
for instance, 2A f 3A are not exactly equivalent to 5A. This
difficulty has been entirely obviated by Dennett's unit of the
centrad, which is the hundredth part of a radian, and is denoted
by a reversed delta (V). Each unit is practically 34 '377', but
the multiples of a centrad are appreciably greater than the mul
tiples of a prism dioptre ; and this is even of an advantage in
using the formula N = ^V, owing to spherical aberration. It
has been adopted as the official unit by the American Ophthal
mological Society, and before long I hope it will be universally
used in this country. When the centrad is used, 3V f 4V are
exactly equivalent to 7V, and 10V is ten times the strength
of IV.
What is the effect of a +25D lens decentred 4 mm. in?
ID (4)(25)
~ 10 V 10
The effect will be that of a normally centred +25D com
bined with a prism of one centrad set edge outwards (i.e. it will
be abducting in function).
FORMULA 127
FORMULA FOR REFERENCE.
A. Universally true
(a) For thin centric pencils
For oblique pencils when ^ and v. 2 carry the same sign
=v as w
(c) Circle of least confusion
f i + *> 2  2s
 sfa +
i 4 Vz  2s
B. Refraction at a Spherical Surface
When p Q is the index of the medium in which the source of
light lies, and // is the index of the refracting medium
 Mo Mo
ECCENTRIC PENCILS
2
sn < 
/AO u fv 2 M r sn
C. Lenses
MAGNIFICATION
M
B
(
When KJ = J = rf + q, (i) becomes ^TT*. (u) becomes
128 APPENDIX
. CARDINAL POINTS
When =+//' / a ',
ft /*"/ ff fiifii
ii /I* in _/ l _ "Vi/g TJH /1/2
h= K> * ~K' F ~~K~' F = ~K'
In complex systems, t = H a "H 6 ', V = H'H a ', h" = H"H 6 ".
DIOPTRES
j QQ
When f given in centimetres D = jr\ when/' in millimetres,
DECENTRATION AND CENTRADS
THIN LENSES, OBLIQUE CENTRIC PENCILS
(ii) When incident rays are parallel
D sin Q 4Q n.
D * ~jT^i ' ~~8i^' ' DC
D. Prisms
D = i/f  <#>  A.
MINIMUM DEVIATION
D = 2,/r  A.
sin }A
If A small, D = p  1A.
In order to convert the formulae (B) into those suitable for a
plane surface, put r = oo . To convert them into the correspond
ing formulae for reflection, put /u, = 1 and <j>' = <j>.
ANSWERS
CHAPTEB I (p. 10).
(1) 45 feet. (2) 70 feet.
(8) The intensity of the electric light is 900 times that of the
gaslight.
(4) The shadow cast by the gasflame will be approximately 8
times more intense than that cast by the electric light.
CHAPTEE II (p. 17).
(1) 1 foot, 5 feet, 7 feet. (2) 7 images.
(3) One image. If ECB < 60, two images would be seen. For
PCB = 120 since AGP = 15 and ACB = 135. Produce BC to c,
and draw Pp 6 perpendicular to BC produced, cutting it at c ; make cp b
= PC ; then the angle p b Cc = 60. Consequently, the line p 6 E will
only cut the mirror when ECB < 60.
CHAPTEB III (p. 33).
(1) The image is 15 cm. in height, it is virtual and erect, and it is
formed 16 cm. behind the mirror.
(2) (i.) 24 inches in front of the mirror,
(ii.) 8 inches behind the mirror.
(3) r =  6 inches ; the mirror is convex.
(4) p = 6^ inches,  =  11 ; the image is inverted.
CHAPTEB IV (p. 48).
(2) M = 154. (3) n = V2.
(5) Two prisms of minimum deviations 1 44' and 2 respectively,
if properly placed, would correct the deviation. The weaker prism
should be placed edge outwards before the left eye, while the 2 prism
should be placed edge outwards and upwards before the right eye, so
that the base apex line makes with the horizontal line an angle of 30.
129
130 ANSWEKS
CHAPTEE V (p. 67).
(1) An inverted image mm. high would be formed 22'2 mm.
from the cornea, i.e. 22 mm. behind the second principal focus.
150 _ 1
~ io' lc ~
I a _ 06 ._ 10
I c ~ 06 ~ 9*
(3) A real image will be formed at the distance of the diameter of
the sphere from the unsilvered side.
CHAPTER VI (p. 107).
= 72 inches.
=' /./'/"=CT 
(3) r 2 39 inches ; in water /'= 37 inches.
(4) (i) M = 485, (ii) M = 435, (iii) M = 50.
(5) h' or H'A = 16, h" or H"B = 0'8, FH' = 32, F"H"= 3'2.
INDEX
ABERRATION, lateral, 120
longitudinal, 31, 120
spherical lens, 8991
reflection, 3033
refraction, 6465
Aplanatism, 31
Astigmatic pencil, 3133
Axial pencil, 18, 68
BADEN POWELL lens, 104105
Brightness, 46
CAMERA lucida, 111
Cardinal Points, 9196
of eye, 99101
Caustic, 3031, 120
Centrads, 126
Centre of lens, 8284
Centric pencils, 18, 68, 83
Circle of least aberration, 120
Circle of least confusion, 117120
Coddington lens, 8687
Conjugate focus, 1920
Conversion of formulae, 6567
Critical angle, 37
DECENTRED lens, 124126
Dense medium, 34
Deviation, prism, 4344
prismosphere, 124
refraction, 36
rotating mirror, 1415
Dioptre, 8081
ECCENTRIC pencils, 3033, 6465,
109110, 111117
Eye, 99102
FOCAL lines reflection, 3033, 109
110
refraction, lens, 121123
plane surface, 111113
plate, 113114
spherical surface, 6465, 114
116
GAUSS, 92
Graphic methods, for image, 2930,
6164, 80
for cardinal points, 105107
HUYGENIAN eyepiece, 9697
ILLUMINATION, 14
Image, lens, 6977
plane reflection, 1317
plane refraction, 3840
prism, 4546
real and virtual, 13
spherical mirror, 1830
spherical surface, 5064
KALEIDOSCOPE, 15
LAWS of illumination, 14
propagation, 12
reflection, 11
refraction, 3436
Lenses, 68107
Light sense, 3
MAGNIFICATION, 7780i
Mirrors, plane, 1114
concave, 1821
convex, 2224
NODAL points, 84, 8789, 100102
OBLIQUE pencils, lens, 121123
plane surface, 111114
spherical surface, 3033, 109110,
114117
Opera glass, 98
Optical centre, 8284
PHOTOMETER, 3
Pinholes, 78
Power, 8081
Principal points, 9196
Prism, 4147
Prism dioptres, 124125
Prismosphere, 124
131
132 INDEX
REFLECTION plane, 1115
repeated, 1517
spherical surface, 1833
Refraction, plane, 3447
Refractive index, 35
Rotation of mirrors, 1415
of prisms, 4647
SHADOWS, 89
Signs, 2122, 49
Size of image, 2627, 5961, 7377
Snell, 34
Sphere, 8486
Stanhope lens, 87
TELESCOPE, penetrating power, 6
Total reflection, 37
Tscherning, 99
ULTEAMICROSCOPE, 6
Umbra, 9
Unit planes, 100
VISUAL angle, 67, 77, 102
WOLLASTON prism, 111
THE END
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